Hey there, future scientists! Welcome to the exciting world of Thermodynamics! Today, we're diving into one of the most fundamental principles in all of physics – the First Law of Thermodynamics. It's super intuitive once you get the hang of it, and it's basically a fancy way of saying "energy can't be created or destroyed, only transformed." You've probably heard that before, right? Well, this law puts that idea into action when we talk about heat, work, and the internal energy of a system.

### 1. The Grand Idea: Conservation of Energy (Again!)

Let's start with a principle you might already be familiar with: the Law of Conservation of Energy. It states that energy can neither be created nor destroyed; it can only be transformed from one form to another or transferred from one system to another. Think about it: when you rub your hands together, the kinetic energy of your hands isn't destroyed; it's converted into heat energy. When a car burns fuel, the chemical potential energy in the fuel isn't destroyed; it's converted into kinetic energy (to move the car) and heat energy (making the engine hot).

The First Law of Thermodynamics is essentially this conservation law applied to a specific context: when we're dealing with heat flow, work done, and the internal energy of a system. It helps us track how energy moves in and out of a system and what happens to the energy stored within it.

### 2. Setting the Stage: System, Surroundings, and Boundary

Before we get into the nitty-gritty, let's quickly recap some basic terms that are crucial for understanding thermodynamics:

* System: This is the part of the universe we are interested in studying. It could be anything – the gas inside a cylinder, a cup of coffee, or even your own body.
* Surroundings: Everything else in the universe outside the system that can interact with it.
* Boundary: The real or imaginary surface that separates the system from its surroundings. Energy and/or matter can cross this boundary depending on the type of system.

For the First Law, we usually focus on systems where energy can be exchanged, but matter cannot (a "closed system"). Imagine a sealed pressure cooker – heat can go in or out, but the food and steam stay inside.

### 3. The Three Musketeers of the First Law: Internal Energy, Heat, and Work

The First Law connects three main concepts. Let's understand each one intuitively:

#### a) Internal Energy (U or E) - The System's Energy Bank Account

Imagine your system has a "bank account" of energy. This is its internal energy (U). It's the total energy contained *within* the system due to the motion and interaction of its molecules.
* What's inside? It includes the kinetic energy of random motion of atoms and molecules (translational, rotational, vibrational) and the potential energy associated with intermolecular forces.
* Temperature's Role: For an ideal gas (which we often assume in basic thermodynamics), internal energy depends *only* on its temperature. If the temperature increases, the molecules move faster, and their internal energy increases. If the temperature decreases, internal energy decreases.
* State Function: This is important! Internal energy is a state function. This means its value depends only on the current state of the system (e.g., its temperature, pressure, volume), *not* on how it got to that state. If you have water at 25°C and 1 atm pressure, its internal energy is fixed, regardless of whether you heated it from 10°C or cooled it from 50°C. We are usually interested in the change in internal energy (ΔU).

#### b) Heat (Q) - Energy Transfer Due to Temperature Difference

Think of heat (Q) as energy *flowing* into or out of your system because there's a temperature difference between the system and its surroundings.
* Hot to Cold: Heat always flows naturally from a region of higher temperature to a region of lower temperature.
* Example: If you put a cold drink in a warm room, heat flows from the warmer room (surroundings) into the colder drink (system), making the drink warmer. If you put a hot cup of coffee on the table, heat flows from the coffee to the cooler room.
* Path Function: Unlike internal energy, heat is a path function. This means the amount of heat transferred depends on *how* the process occurred, not just the initial and final states.

#### c) Work (W) - Energy Transfer Not Due to Temperature Difference

Work (W) is another way energy can be transferred into or out of your system, but *not* because of a temperature difference. In thermodynamics, we often talk about pressure-volume work (or P-V work), which is done when a system expands or contracts against an external pressure.
* Example: Imagine gas trapped in a cylinder with a movable piston.
* If the gas expands, it pushes the piston outwards, doing work *on* the surroundings. The system loses energy.
* If the piston pushes inwards, compressing the gas, the surroundings do work *on* the system. The system gains energy.
* Other Forms of Work: While P-V work is common, work can also be electrical (e.g., current flowing through a resistor in the system) or mechanical (e.g., stirring a liquid).
* Path Function: Just like heat, work is also a path function. The amount of work done depends on the specific process or path taken.

### 4. The First Law of Thermodynamics: The Equation!

Now, let's put it all together! The First Law of Thermodynamics states:

ΔU = Q + W

Where:
* ΔU is the change in the internal energy of the system.
* Q is the heat added *to* the system.
* W is the work done *on* the system.

This equation beautifully captures the conservation of energy. It says that any change in the system's internal energy (ΔU) must be due to either heat flowing into/out of the system (Q) or work being done on/by the system (W).

#### Understanding the Sign Conventions (This is CRUCIAL for JEE!)

The signs of Q and W are super important to get right. Here's the most common convention used in JEE and many textbooks:

Quantity	Positive (+) Sign	Negative (-) Sign
Heat (Q)	Heat is ADDED TO the system (system gains energy)	Heat is REMOVED FROM the system (system loses energy)
Work (W)	Work is DONE ON the system by the surroundings (system gains energy, e.g., compression)	Work is DONE BY the system on the surroundings (system loses energy, e.g., expansion)
Change in Internal Energy (ΔU)	Internal energy INCREASES (temperature often increases for ideal gases)	Internal energy DECREASES (temperature often decreases for ideal gases)

Analogy: Think of your personal bank account.
* Q is like a direct deposit: If someone sends money *to* your account, your balance goes up (Q is positive). If you send money *out* of your account, your balance goes down (Q is negative).
* W is like someone else putting money in for you (work ON you): If a friend pays your bill, your effective balance increases (W is positive). If you pay someone else's bill (work *BY* you), your balance decreases (W is negative).
* ΔU is your change in bank balance: If money comes in (Q or W), your balance goes up. If money goes out, your balance goes down. It's that simple!

Important Note for JEE: Sometimes, especially in older textbooks or specific engineering contexts, you might see the First Law written as ΔU = Q - W', where W' is the work done *by* the system. In this case, if the system does work (expansion), W' would be positive, leading to -W' (loss of internal energy). However, for JEE, generally stick to ΔU = Q + W, where W is work done *on* the system. This ensures consistency with the sign conventions mentioned above.

### 5. Applying the First Law: Basic Thermodynamic Processes

The First Law becomes incredibly useful when we look at specific types of processes where one of the variables (P, V, T, Q, or W) is kept constant or zero.

Let's assume we have an ideal gas in a cylinder with a piston, and we're talking about P-V work. The work done *on* the system during a quasi-static process is given by:

W = - ∫ P_external dV

If the external pressure is constant and equal to the system pressure, then:

W = - PΔV = - P(V_final - V_initial)

Remember, this is work done *on* the system.
* If expansion occurs (V_final > V_initial, so ΔV is positive), then W will be negative, meaning the system *did work*.
* If compression occurs (V_final < V_initial, so ΔV is negative), then W will be positive, meaning work was *done on* the system.

Now, let's explore some common processes:

#### a) Isochoric Process (Constant Volume)

* What happens? The volume of the system remains constant (ΔV = 0). Imagine heating gas in a rigid, sealed container.
* Work Done (W): Since ΔV = 0, the work done W = -PΔV = 0. No work is done!
* First Law: ΔU = Q + 0 => ΔU = Q
* Meaning: Any heat added to an isochoric system goes entirely into increasing its internal energy (and thus its temperature). If heat is removed, internal energy decreases.

#### b) Isobaric Process (Constant Pressure)

* What happens? The pressure of the system remains constant. Think of a gas in a cylinder with a piston that can move freely against a constant external atmospheric pressure.
* Work Done (W): Since pressure is constant, W = -PΔV.
* First Law: ΔU = Q - PΔV
* Meaning: In an isobaric process, heat added (Q) can both change the internal energy (ΔU) and do work (PΔV). For example, if you heat a gas at constant pressure, it will expand, doing work on the surroundings, *and* its temperature (internal energy) will increase.

#### c) Isothermal Process (Constant Temperature)

* What happens? The temperature of the system remains constant (ΔT = 0). This usually requires the system to be in good thermal contact with a large reservoir at constant temperature.
* Internal Energy (ΔU): For an ideal gas, internal energy depends *only* on temperature. So, if ΔT = 0, then ΔU = 0.
* First Law: 0 = Q + W => Q = -W
* Meaning: Any heat added to an isothermal ideal gas system must be exactly balanced by work done *by* the system (expansion). Conversely, if work is done *on* the system (compression), the heat generated must be released to the surroundings to keep the temperature constant.

#### d) Adiabatic Process (No Heat Exchange)

* What happens? No heat is allowed to enter or leave the system (Q = 0). This happens very quickly, or in a well-insulated system.
* First Law: ΔU = 0 + W => ΔU = W
* Meaning: If you do work *on* an adiabatically isolated system (e.g., rapidly compress a gas), its internal energy (and temperature) will increase. If the system does work *by expanding*, its internal energy (and temperature) will decrease. This is why a bicycle pump gets hot when you pump air quickly (work done on the gas, no time for heat to escape), and why gases coming out of a spray can feel cold (gas expands rapidly, doing work, internal energy drops).

### 6. Example Time!

Let's put this into practice with a quick example:

Problem: A gas in a cylinder absorbs 200 J of heat from its surroundings and simultaneously expands, doing 80 J of work on the surroundings. Calculate the change in internal energy of the gas.

Step-by-step Solution:

1. Identify the given values and their signs:
* Heat absorbed *by* the system: Q = +200 J (positive because heat is added *to* the system).
* Work done *by* the system *on* the surroundings: W = -80 J (negative because the system is doing the work, losing energy).

2. State the First Law of Thermodynamics:
* ΔU = Q + W

3. Substitute the values:
* ΔU = (+200 J) + (-80 J)
* ΔU = 200 J - 80 J

4. Calculate ΔU:
* ΔU = 120 J

Conclusion: The internal energy of the gas increased by 120 J. This makes sense: the system gained more energy through heat than it lost through work done, so its overall internal energy went up.

### 7. CBSE vs. JEE Focus

* CBSE: For CBSE, understanding the statement of the First Law, the sign conventions, and the basic applications to isochoric, isobaric, isothermal, and adiabatic processes is key. Simple numerical problems like the one above are common.
* JEE: For JEE, you'll need a much deeper understanding. You'll encounter more complex problems involving calculating work done for various non-linear paths on P-V diagrams, cyclic processes, and combining different processes. The subtleties of path functions vs. state functions become critical, and derivations for work done in specific processes (e.g., isothermal reversible expansion of an ideal gas) are essential. Expect problems that require integrating P dV and relating internal energy changes to specific heat capacities.

The First Law is the bedrock of thermodynamics. Master these fundamentals, and you'll be well on your way to tackling more advanced concepts in thermodynamics! Keep practicing those sign conventions – they're your best friend (or worst enemy!) in this topic!

Alright class, let's embark on a deep dive into one of the most fundamental and powerful principles in physics: the First Law of Thermodynamics. This law is essentially a restatement of the principle of conservation of energy, specifically applied to thermodynamic systems. It quantifies how energy is transferred and transformed within a system. For JEE aspirants, a thorough understanding of this law, its components, and its application to various processes is absolutely critical.

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### 1. The First Law of Thermodynamics: A Statement of Energy Conservation

At its heart, the First Law of Thermodynamics tells us that energy cannot be created or destroyed; it can only be converted from one form to another. When applied to a thermodynamic system, this means that any change in the system's internal energy must be accounted for by the heat exchanged with its surroundings and the work done by or on the system.

The mathematical statement of the First Law is:

ΔU = Q + W

Where:
* ΔU represents the change in the internal energy of the system.
* Q represents the heat exchanged with the surroundings.
* W represents the work done (by or on the system).

Let's dissect each of these terms, as their precise understanding is paramount.

#### 1.1 Internal Energy (ΔU)

The internal energy (U) of a system is the sum of the kinetic and potential energies of its constituent particles (atoms and molecules).
* Kinetic energy: Due to the translational, rotational, and vibrational motions of molecules.
* Potential energy: Due to the intermolecular forces between molecules.

For an ideal gas, intermolecular forces are negligible, so the internal energy depends *only* on the kinetic energy of its molecules, which in turn depends *only* on its absolute temperature (T). This is a crucial concept for JEE problems involving ideal gases.

Key Point for Ideal Gases:
If the temperature of an ideal gas does not change (ΔT = 0), then its internal energy does not change (ΔU = 0).
* For a monatomic ideal gas, U = (3/2)nRT
* For a diatomic ideal gas, U = (5/2)nRT (at moderate temperatures)
* In general, for any ideal gas, ΔU = nC_vΔT, where C_v is the molar specific heat at constant volume.

Nature of Internal Energy: Internal energy (U) is a state function. This means its value depends only on the initial and final states of the system (e.g., P, V, T), not on the path taken to reach those states. Therefore, ΔU depends only on U_final - U_initial.

#### 1.2 Heat (Q)

Heat (Q) is the energy transferred between a system and its surroundings due to a temperature difference. It is a form of energy transfer, not a form of energy stored within the system.

Nature of Heat: Heat (Q) is a path function. The amount of heat transferred depends on the specific process (path) followed between the initial and final states.

#### 1.3 Work (W)

Work (W) in thermodynamics typically refers to mechanical work, specifically pressure-volume (P-V) work. This occurs when a system expands or contracts against an external pressure.

Consider a gas confined in a cylinder with a movable piston. If the gas expands, it pushes the piston outwards, doing work on the surroundings. If the surroundings push the piston inwards, work is done on the gas.

For an infinitesimal change in volume dV against an external pressure P_ext, the work done by the system is dW = P_extdV.
For a finite change, W = ∫ P_extdV from V_initial to V_final.

Nature of Work: Work (W) is also a path function. The amount of work done depends on the specific process (path) followed between the initial and final states. On a P-V diagram, the work done is represented by the area under the P-V curve.

#### 1.4 Sign Conventions for Q and W (CRUCIAL for JEE)

Understanding the sign conventions is absolutely vital for correctly applying the First Law in problem-solving. While different conventions exist, the most common and widely accepted for JEE is:

Term	Positive (+)	Negative (-)
Q (Heat)	Heat absorbed by the system (Endothermic)	Heat released by the system (Exothermic)
W (Work)	Work done on the system (Compression)	Work done by the system (Expansion)
ΔU (Internal Energy)	Internal energy increases (Temperature increases for ideal gas)	Internal energy decreases (Temperature decreases for ideal gas)

Important Note: Some textbooks/sources use the convention W = -∫PdV, meaning W is work done *by* the system. In that case, the First Law becomes ΔU = Q - W_{by system}. However, the convention ΔU = Q + W_{on system} (where W is work done *on* the system) is more prevalent in many JEE contexts and simplifies calculations. We will stick to ΔU = Q + W where W is work done *on* the system.

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### 2. Applications of the First Law to Various Thermodynamic Processes

The First Law finds its most practical application when analyzing specific thermodynamic processes. Let's look at the four primary processes:

#### 2.1 Isochoric Process (Constant Volume, ΔV = 0)

In an isochoric process, the volume of the system remains constant.
* Since ΔV = 0, the work done (W = ∫PdV) is zero.
* Applying the First Law: ΔU = Q + W becomes ΔU = Q.
* This means all heat added to the system goes directly to increasing its internal energy (and thus its temperature for an ideal gas).
* For an ideal gas, Q = nC_vΔT. So, ΔU = nC_vΔT.

#### 2.2 Isobaric Process (Constant Pressure, ΔP = 0)

In an isobaric process, the pressure of the system remains constant.
* Work done: W = -P_extΔV. If P_ext = P_system (reversible process), then W = -PΔV. (Remember, this W is work done *by* the system. If we stick to work *on* the system, W = -PΔV implies work done *by* system is PΔV, so work *on* system is -PΔV).
* Let's use the convention: W is work done *on* the system.
* If gas expands (ΔV > 0), work done *by* system is PΔV (positive). Work done *on* system is W = -PΔV (negative).
* If gas compresses (ΔV < 0), work done *by* system is PΔV (negative). Work done *on* system is W = -PΔV (positive).
* Applying the First Law: ΔU = Q - PΔV. (This is if W = -PΔV is taken as work *by* system. If we strictly adhere to W as work *on* system, it's ΔU = Q + W, where W = -PΔV for expansion).
* For an ideal gas, Q = nC_pΔT. So, ΔU = nC_pΔT - PΔV.
* Also, remember that C_p - C_v = R (Mayer's relation).

#### 2.3 Isothermal Process (Constant Temperature, ΔT = 0)

In an isothermal process, the temperature of the system remains constant.
* For an ideal gas, since ΔT = 0, the change in internal energy is ΔU = 0.
* Applying the First Law: ΔU = Q + W becomes 0 = Q + W, or Q = -W.
* This means any heat absorbed by the system is entirely converted into work done *by* the system, and vice-versa.
* Work done for a reversible isothermal process (ideal gas):
* W = -∫PdV (work done *by* the system)
* From ideal gas law, P = nRT/V.
* W = -∫(nRT/V)dV = -nRT ∫(1/V)dV from V₁ to V₂
* W_{by system} = -nRT ln(V₂/V₁) = -nRT ln(P₁/P₂)
* If we define W as work *on* the system, then W_{on system} = nRT ln(V₂/V₁) = nRT ln(P₁/P₂). (Note the change in sign and the log terms are flipped depending on which V/P is in numerator).
* Let's maintain consistency: if W is work *on* the system, W = - ∫P dV (where P is external, and for reversible P_ext = P_system).
* So, W = - nRT ln(V_f/V_i).
* Then Q = -W = nRT ln(V_f/V_i).

#### 2.4 Adiabatic Process (No Heat Exchange, Q = 0)

In an adiabatic process, no heat is exchanged between the system and its surroundings. This can occur if the system is perfectly insulated or if the process happens very rapidly.
* Since Q = 0, the First Law becomes ΔU = W.
* This means any work done on the system increases its internal energy (and temperature), and any work done by the system decreases its internal energy (and temperature).
* For a reversible adiabatic process (ideal gas), the following relations hold:
* PV^γ = constant
* TV^γ-1 = constant
* TP^(1-γ)/γ = constant or T^γP^1-γ = constant
Where γ (gamma) = C_p/C_v is the adiabatic index or ratio of specific heats.

* Work done for a reversible adiabatic process (ideal gas):
* Since Q=0, ΔU = W.
* We know ΔU = nC_vΔT = nC_v(T_f - T_i).
* So, W = nC_v(T_f - T_i).
* Using C_v = R/(γ-1), we get W = nR(T_f - T_i)/(γ-1).
* Also, W = (P_fV_f - P_iV_i)/(γ-1) (using PV=nRT).
* Again, this W is work done *on* the system. If work *by* system is asked, reverse the sign.

#### 2.5 Cyclic Process

A cyclic process is a series of changes that returns the system to its initial state.
* Since the initial and final states are the same, the change in internal energy is ΔU = 0.
* Applying the First Law: ΔU = Q + W becomes 0 = Q + W, or Q = -W.
* This means the net heat absorbed by the system equals the net work done *by* the system over the cycle.
* On a P-V diagram, a cyclic process is represented by a closed loop. The area enclosed by the loop represents the net work done during the cycle.
* If the cycle is traversed clockwise, W_net is negative (work done *by* system is positive).
* If the cycle is traversed counter-clockwise, W_net is positive (work done *on* system is positive).

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### 3. JEE Advanced Perspective: P-V Diagrams and Work Done

P-V diagrams are incredibly powerful tools for visualizing thermodynamic processes and calculating work done.

* The area under the curve on a P-V diagram represents the magnitude of the work done during a process.
* Expansion: If volume increases (moving right on the P-V diagram), the system does positive work, so work *on* the system is negative.
* Compression: If volume decreases (moving left on the P-V diagram), work is done *on* the system, so work *on* the system is positive.

Calculating Work on P-V Diagrams:
* For a path from (V₁, P₁) to (V₂, P₂), the work done is ∫PdV.
* For complex paths, break them into simpler geometric shapes (rectangles, triangles, trapezoids).
* For a cyclic process, the net work done is the area enclosed by the loop. If the loop is clockwise, net work *by* the system is positive (W_{on system} is negative). If counter-clockwise, net work *on* the system is positive (W_{by system} is negative).

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### 4. Illustrative Examples

Let's solidify our understanding with a couple of examples.

Example 1: Isobaric Expansion

An ideal gas expands from 1.0 m³ to 3.0 m³ at a constant pressure of 2.0 x 10⁵ Pa. During this process, the system absorbs 500 kJ of heat.
(a) Calculate the work done by the gas.
(b) Calculate the change in internal energy of the gas.

Solution:

Given:
P = 2.0 x 10⁵ Pa (constant)
V_i = 1.0 m³
V_f = 3.0 m³
Q = +500 kJ = +500 x 10³ J (heat absorbed by system is positive)

(a) Work done *by* the gas (W_by):
For an isobaric process, W_by = PΔV
W_by = P(V_f - V_i)
W_by = (2.0 x 10⁵ Pa) * (3.0 m³ - 1.0 m³)
W_by = (2.0 x 10⁵ Pa) * (2.0 m³)
W_by = 4.0 x 10⁵ J = 400 kJ

If we stick to our convention of W as work *on* the system:
W = -W_by = -400 kJ.

(b) Change in internal energy (ΔU):
Using the First Law: ΔU = Q + W
ΔU = (+500 kJ) + (-400 kJ)
ΔU = 100 kJ

The internal energy of the gas increased by 100 kJ.

Example 2: Isothermal Compression

Two moles of an ideal gas are compressed isothermally and reversibly from 10 L to 2 L at a temperature of 300 K.
(a) Calculate the work done on the gas.
(b) Calculate the heat exchanged with the surroundings.
(c) What is the change in internal energy of the gas? (Given R = 8.314 J mol^-1 K^-1)

Solution:

Given:
n = 2 moles
V_i = 10 L = 10 x 10^-3 m³
V_f = 2 L = 2 x 10^-3 m³
T = 300 K (constant)
R = 8.314 J mol^-1 K^-1

(a) Work done *on* the gas (W):
For a reversible isothermal process, W = nRT ln(V_f/V_i)
W = (2 mol) * (8.314 J mol^-1 K^-1) * (300 K) * ln(2 L / 10 L)
W = 2 * 8.314 * 300 * ln(0.2)
W = 4988.4 * (-1.6094)
W ≈ -8028 J (This is work done *on* the system, so the negative sign implies work is done *by* the system. Let's re-evaluate using the work done *by* the system formula first, then convert.)

Let's calculate work done *by* the system: W_by = -nRT ln(V_f/V_i)
W_by = - (2 mol) * (8.314 J mol^-1 K^-1) * (300 K) * ln(2 L / 10 L)
W_by = - (4988.4) * (-1.6094)
W_by ≈ +8028 J

So, work done *on* the gas (W) = -W_by = -8028 J.
The positive value of W_by makes sense, as compression means work is done *on* the gas by external forces, but the formula -nRT ln(Vf/Vi) is work done *by* the system.
Let's just use the direct formula for work *on* the system: W = ∫ P dV.
Since P = nRT/V, W = ∫ (nRT/V) dV from V_i to V_f.
W = nRT [ln V] from V_i to V_f = nRT (ln V_f - ln V_i) = nRT ln(V_f/V_i).
W = (2)(8.314)(300) ln(2/10) = 4988.4 * ln(0.2) = 4988.4 * (-1.6094) = -8028 J.
Ah, this convention for W = nRT ln(Vf/Vi) represents the work done *by* the surroundings *on* the system.
Since V_f < V_i, ln(V_f/V_i) will be negative, making W negative. This means the system does work *on* the surroundings.
This is where sign convention can be tricky! Let's clarify:
If we define W as work done *on* the system:
W = nRT ln(V_f/V_i)
In this case, W = (2)(8.314)(300) ln(2/10) = -8028 J.
Since W is negative, it means work is done *by* the system. But it's compression, so work should be done *on* the system.
This indicates my initial formula for W (work on system) was likely for expansion when W = -P_ext dV is used and then inverted.

Let's be crystal clear with the most common JEE approach:
Work done *by* the system: W_by = ∫ P dV
For isothermal reversible: W_by = nRT ln(V_f/V_i)
In our case: W_by = (2)(8.314)(300) ln(2/10) = -8028 J.
Since W_by is negative, it means work is done *on* the system.
So, work done *on* the gas is +8028 J.
Thus, in the First Law ΔU = Q + W, we use W = +8028 J.

(b) Change in internal energy (ΔU):
For an ideal gas in an isothermal process, ΔT = 0, therefore ΔU = 0.

(c) Heat exchanged (Q):
Using the First Law: ΔU = Q + W
0 = Q + (+8028 J)
Q = -8028 J
The negative sign means the gas releases 8028 J of heat to the surroundings. This makes sense: during compression, work is done on the gas, increasing its energy. To keep the temperature constant (isothermal), this energy must be released as heat.

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This deep dive should provide you with a robust framework for understanding and applying the First Law of Thermodynamics across various processes. Remember to practice rigorously with P-V diagrams and pay close attention to the sign conventions for Q and W!

Mastering the First Law of Thermodynamics is crucial for both JEE Main and board exams. Here are some effective mnemonics and shortcuts to help you remember the key concepts and their applications.

1. Mnemonic for the First Law Equation

The First Law of Thermodynamics is stated as: $Delta U = Q - W$ (Physics convention, where W is work done BY the system).

• "You Are Q-W": This simple phrase directly translates to the equation.
  
  
  
  • U: $Delta U$ (Change in Internal Energy) - "You"
  
  
  • Are: = (equals)
  
  
  • Q-W: $Q - W$ (Heat added minus Work done by the system)
  
  
  
  

2. Mnemonics for Sign Conventions

Getting the signs right is critical for numerical problems. Remember these:

• For Q (Heat):
  
  
  
  • "Heat IN, POSITIVE WIN!": If heat is absorbed by the system (system gains heat), Q is positive (+ve).
  
  
  • "Heat OUT, NEGATIVE DOUBT!": If heat is released by the system (system loses heat), Q is negative (-ve).
  
  
  
  


• For W (Work Done): (Physics convention: Work done BY the system)
  
  
  
  • "Work BY (system), POSITIVE SKY!": If the system expands and does work on the surroundings, W is positive (+ve).
  
  
  • "Work ON (system), NEGATIVE GONE!": If the surroundings do work on the system (compression), W is negative (-ve).
  
  
  • JEE & CBSE Tip: Be very careful with sign conventions. While physics generally uses $Delta U = Q - W_{by}$, some chemistry texts might use $Delta U = Q + W_{on}$ (where $W_{on}$ is work done ON the system). Always stick to the convention given or implied in your physics syllabus.
  
  
  
  


• For $Delta U$ (Change in Internal Energy):
  
  
  
  • "Temp UP, Energy UP!": If the temperature of the system increases, its internal energy increases, so $Delta U$ is positive (+ve).
  
  
  • "Temp DOWN, Energy DOWN!": If the temperature decreases, its internal energy decreases, so $Delta U$ is negative (-ve).
  
  
  
  

3. Process-Specific Shortcuts and Mnemonics

Applying the First Law to different thermodynamic processes simplifies the equation considerably.

• Isothermal Process ($Delta T = 0$):
  
  
  
  • "Isothermal: T-Zero, U-Zero, Q equals W!"
  
  
  • Since temperature (T) is constant, for an ideal gas, the change in internal energy ($Delta U$) is zero. Thus, the First Law simplifies to $Q = W$.
  
  
  
  


• Adiabatic Process ($Q = 0$):
  
  
  
  • "A-diabatic: A-bsent Q, so U is -W."
  
  
  • There is no heat exchange with the surroundings ($Q=0$). The First Law becomes $Delta U = -W$.
  
  
  
  


• Isochoric Process ($Delta V = 0$):
  
  
  
  • "Isochoric: V-Fixed, W-Zero, U equals Q!"
  
  
  • The volume (V) is constant, meaning no work is done by or on the system ($W = PDelta V = 0$). The First Law simplifies to $Delta U = Q$.
  
  
  
  


• Isobaric Process ($Delta P = ext{constant}$):
  
  
  
  • "Isobaric: Pressure Constant, Work is P Delta V."
  
  
  • The pressure (P) remains constant. Work done is $W = P(V_f - V_i) = PDelta V$. The full First Law equation, $Delta U = Q - PDelta V$, must be used.
  
  
  
  

Quick Recap Table (Short-cut for Processes)

This table summarizes the key characteristics and simplified First Law for each process:

Process	Constant Variable	Implication on First Law ($Delta U = Q - W$)	Mnemonic / Shortcut
Isothermal	Temperature (T)	$Delta U = 0 implies Q = W$	T-Zero, U-Zero, Q=W
Adiabatic	No Heat (Q)	$Q = 0 implies Delta U = -W$	A-bsent Q, U is -W
Isochoric	Volume (V)	$W = 0 implies Delta U = Q$	V-Fixed, W-Zero, U=Q
Isobaric	Pressure (P)	$W = PDelta V$ (use full law)	P-Constant, Work is P Delta V

By remembering these mnemonics and shortcuts, you can quickly recall the crucial aspects of the First Law of Thermodynamics and its applications, saving valuable time in exams.

Mastering the First Law of Thermodynamics is crucial for both JEE Main and board exams. It's the cornerstone of understanding energy transformations in physical and chemical processes. Here are some quick tips to help you ace related problems:

Quick Tips: First Law of Thermodynamics

The First Law of Thermodynamics is essentially the principle of conservation of energy applied to thermodynamic systems.

• Mathematical Form: The most common form used in JEE and CBSE is:
  
  
  $Delta U = Q - W$
  
  
  Where:
  
  
  
  • $Delta U$ = Change in internal energy of the system
  
  
  • $Q$ = Heat supplied to the system
  
  
  • $W$ = Work done by the system
  
  
  
  Note: Some textbooks might use $Delta U = Q + W'$, where $W'$ is work done on the system. Always be consistent with the sign convention you adopt. For JEE, $Delta U = Q - W$ with $W$ as work done by the system is standard.
  



• Crucial Sign Conventions: This is where most students make mistakes!
  
  
  
  • Heat ($Q$):
    
    
    
    • $Q > 0$: Heat absorbed by the system (system gains heat).
    
    
    • $Q < 0$: Heat released by the system (system loses heat).
    
    
    
    
  
  
  • Work ($W$):
    
    
    
    • $W > 0$: Work done by the system (expansion).
    
    
    • $W < 0$: Work done on the system (compression).
    
    
    
    
  
  
  • Internal Energy ($Delta U$):
    
    
    
    • $Delta U > 0$: Internal energy increases (temperature rises for an ideal gas).
    
    
    • $Delta U < 0$: Internal energy decreases (temperature falls for an ideal gas).
    
    
    
    
  
  
  
  



• Internal Energy ($Delta U$): A State Function
  
  
  
  • $Delta U$ depends only on the initial and final states of the system, not on the path taken.
  
  
  • For an ideal gas, $Delta U = nC_v Delta T$, where $n$ is moles, $C_v$ is molar specific heat at constant volume, and $Delta T$ is the change in temperature.
    
  • For a cyclic process, where the system returns to its initial state, $Delta U = 0$.
  
  
  
  



• Work Done ($W$): A Path Function
  
  
  
  • $W = int P dV$. Work done depends on the path taken between initial and final states.
  
  
  • On a P-V diagram, work done is the area under the P-V curve.
    
    
    
    • For expansion (increasing volume), work done by the system is positive.
    
    
    • For compression (decreasing volume), work done by the system is negative.
    
    
    • For a cyclic process, the net work done is the area enclosed by the cycle on the P-V diagram. Clockwise cycle implies positive net work (output), anti-clockwise implies negative net work (input).
    
    
    
    
  
  
  
  

Applications in Specific Processes

Understanding how the First Law simplifies for different processes is key:

Process	Defining Condition	Key Implication for First Law	Work Done (Ideal Gas)
Isochoric	$V = ext{constant}$ ($Delta V = 0$)	$W = 0 implies Delta U = Q$	$W = 0$
Isobaric	$P = ext{constant}$	$Delta U = Q - PDelta V$	$W = PDelta V = P(V_f - V_i)$
Isothermal	$T = ext{constant}$ ($Delta T = 0$)	$Delta U = 0 implies Q = W$	$W = nRT lnleft(frac{V_f}{V_i} ight)$
Adiabatic	$Q = 0$ (no heat exchange)	$Delta U = -W$	$W = frac{nR(T_i - T_f)}{gamma - 1}$ or $W = frac{P_iV_i - P_fV_f}{gamma - 1}$
Cyclic Process	Returns to initial state	$Delta U = 0 implies Q = W$	Area enclosed by the cycle on P-V diagram

Stay sharp with your sign conventions and formula applications. Practice drawing P-V diagrams for different processes to visualize work done. Good luck!

Real-World Applications of the First Law of Thermodynamics

The First Law of Thermodynamics, essentially a statement of the conservation of energy, is one of the most fundamental principles in physics and finds ubiquitous applications in various fields of science and engineering. Understanding its real-world implications is crucial for both theoretical comprehension and practical problem-solving in JEE and Board exams.

Quick Recap: The First Law states that the change in internal energy (ΔU) of a system is equal to the heat (Q) added to the system minus the work (W) done *by* the system: ΔU = Q - W.

Here are some key real-world applications:

• 
  Heat Engines and Power Plants:
  
  
  
  • Internal combustion engines (cars, motorcycles): Fuel combustion releases heat (Q), which increases the internal energy of gases, leading to expansion and work done (W) on pistons. The remaining energy is expelled as exhaust heat. The net work output is directly linked to the difference between heat supplied and heat rejected, consistent with ΔU = Q - W (over a cycle, ΔU ≈ 0).
  
  
  • Thermal power plants: Heat generated from burning fossil fuels or nuclear reactions converts water into high-pressure steam. This steam drives turbines, producing mechanical work (W) that generates electricity.
  
  JEE Focus: Questions often involve calculating efficiency, work done, or heat rejected in ideal engine cycles (e.g., Carnot, Otto, Diesel), directly applying the First Law for each stage of the cycle.
  
  
  


• 
  Refrigerators and Air Conditioners (Heat Pumps):
  
  
  
  • These devices work in reverse of heat engines. They extract heat (Q) from a cold reservoir (e.g., inside a fridge or a room) and reject it to a warmer one (the surroundings) by doing external work (W) on the refrigerant. This process also adheres to ΔU = Q - W, where Q is the heat absorbed and W is the work input.
  
  
  
  


• 
  Human Body and Metabolism:
  
  
  
  • The human body is an excellent example of an open thermodynamic system. Food consumed (chemical energy) is converted into heat (maintaining body temperature, Q) and mechanical work (muscle movement, W). The change in internal energy (ΔU) of the body corresponds to the energy stored or released. This is a direct application of energy conservation in biological systems.
  
  
  • CBSE Focus: Often tested conceptually, relating energy intake from food to energy expenditure and heat generation.
  
  
  
  


• 
  Weather and Atmospheric Processes:
  
  
  
  • The formation of clouds, rain, and winds involves thermodynamic processes. For instance, adiabatic expansion of rising air parcels (Q ≈ 0) causes cooling and condensation (ΔU = -W). Similarly, atmospheric convection involves heat transfer and work done by air masses.
  
  
  
  


• 
  Chemical Reactions (Calorimetry):
  
  
  
  • In chemical reactions, the change in internal energy (ΔU) or enthalpy (ΔH) is measured using calorimeters. These measurements quantify the heat released or absorbed (Q) and the work done (W) during the reaction, providing vital information for energy analysis in chemistry.
  
  
  
  

Understanding these real-world examples solidifies your grasp of the First Law, making it easier to tackle complex problems. Always remember that energy is conserved; it merely transforms from one form to another.

Related Topics to the First Law of Thermodynamics

Understanding the First Law of Thermodynamics requires a grasp of several foundational concepts and naturally leads to the study of other interconnected topics. For success in both board exams and JEE, it's crucial to see how these concepts fit together, forming a comprehensive understanding of thermodynamics.

• Zeroth Law of Thermodynamics:
  
  
  
  • Connection: While the First Law deals with energy conservation, the Zeroth Law establishes the concept of temperature and thermal equilibrium. Temperature is a key state variable, essential for defining heat flow (Q) and internal energy (U), both central to the First Law. Without a clear understanding of temperature, the concepts of "heat" and "thermal contact" lack rigorous definition.
  
  
  
  


• Internal Energy (U):
  
  
  
  • Connection: This is a core state function whose change is directly governed by the First Law (ΔU = Q - W). A deep understanding of internal energy, including its dependence on temperature for ideal gases (U = nC_vT or U = f/2 nRT, where f is degrees of freedom), is indispensable for applying the First Law to various processes.
  
  
  
  


• Work Done (W) by/on a Gas:
  
  
  
  • Connection: The First Law explicitly involves work (W) as a mechanism for energy transfer. Understanding how to calculate work done in different thermodynamic processes (isobaric, isochoric, isothermal, adiabatic) is paramount for applying the First Law. The formula W = ∫PdV is fundamental here.
  
  
  
  


• Heat (Q) and Specific Heat Capacities:
  
  
  
  • Connection: Heat (Q) is the other mode of energy transfer in the First Law. Concepts like specific heat capacity at constant volume (C_v) and constant pressure (C_p) are directly related to the First Law. Q = nCΔT is used to calculate heat, and the relationship C_p - C_v = R (Mayer's relation), which is derivable from the First Law for ideal gases, is crucial for problem-solving.
  
  
  
  


• Thermodynamic Processes (Isothermal, Adiabatic, Isobaric, Isochoric):
  
  
  
  • Connection: These specific processes are the primary applications of the First Law. Each process involves specific conditions (e.g., constant temperature, no heat exchange, constant pressure, constant volume) that simplify the First Law equation (ΔU = Q - W), allowing for the calculation of Q, W, and ΔU under those conditions. Mastering these applications is critical for JEE problems.
  
  
  
  


• Kinetic Theory of Gases (KTG):
  
  
  
  • Connection: KTG provides the microscopic foundation for macroscopic thermodynamic quantities like internal energy, pressure, and temperature. Understanding the relationship between the average kinetic energy of gas molecules and temperature, and how this relates to internal energy, strengthens the conceptual basis for the First Law.
  
  
  
  


• Second Law of Thermodynamics:
  
  
  
  • Connection: While the First Law addresses the conservation of energy, the Second Law introduces the concept of entropy and dictates the directionality of spontaneous processes and the limits of energy conversion efficiency. It builds upon the First Law by providing additional constraints on what processes are possible, even if energetically allowed by the First Law.
  
  
  
  

JEE/CBSE Perspective: For JEE, you'll often encounter problems that combine concepts from the First Law with KTG (e.g., calculating degrees of freedom or molar specific heats from internal energy expressions). For CBSE, the focus remains on understanding the definitions and applying the First Law to the standard thermodynamic processes.

By studying these related topics alongside the First Law, you build a robust and interconnected understanding of thermodynamics, essential for excelling in exams.

Solving problems involving the First Law of Thermodynamics requires a systematic approach, especially given the various thermodynamic processes and their associated formulas. Mastering the sign conventions is paramount to avoiding common errors.

Systematic Problem-Solving Approach for First Law of Thermodynamics

The First Law of Thermodynamics is essentially the law of conservation of energy applied to thermodynamic systems. Its mathematical form is generally expressed as:

$Delta U = Q + W$

• $Delta U$: Change in internal energy of the system.


• $Q$: Heat added to the system.


• $W$: Work done on the system.

Key Steps:

1. 
   Define the System Clearly:
   
   
   
   • Identify what constitutes the 'system' (e.g., a gas in a cylinder, a block of ice, etc.). This helps in tracking heat and work interactions correctly.
   
   
   
   


2. 
   Identify the Thermodynamic Process:
   
   
   
   • Determine if the process is isochoric (constant volume, $W=0$), isobaric (constant pressure), isothermal (constant temperature, $Delta U=0$ for ideal gas), adiabatic ($Q=0$), or cyclic ($Delta U=0$ over a complete cycle).
   
   
   • This identification dictates which terms in the First Law equation might be zero or how they relate to each other.
   
   
   
   


3. 
   Master Sign Conventions: (This is the most critical step and a frequent source of error in JEE/CBSE.)
   
   
   
   • Heat (Q):
     
     
     
     • $Q > 0$: Heat absorbed by the system.
     
     
     • $Q < 0$: Heat rejected by the system (or absorbed by surroundings).
     
     
     
     
   
   
   • Work (W): (Using the convention $W$ = work done on the system)
     
     
     
     • $W > 0$: Work done on the system (e.g., compression). Volume decreases.
     
     
     • $W < 0$: Work done by the system (e.g., expansion). Volume increases.
     
     
     
     

     
     Note for JEE: Some resources use $W'$ for work done *by* the system, where $Delta U = Q - W'$. Be consistent with your chosen convention. Here, we use $W$ for work done *on* the system.
     

     
     
   
   
   • Change in Internal Energy ($Delta U$):
     
     
     
     • $Delta U > 0$: Internal energy increases (temperature increases for ideal gas).
     
     
     • $Delta U < 0$: Internal energy decreases (temperature decreases for ideal gas).
     
     
     
     
   
   
   
   


4. 
   List Relevant Formulas:
   
   
   
   • Internal Energy for Ideal Gas: $Delta U = n C_v Delta T$.
     
     
     
     • For monoatomic gas: $C_v = frac{3}{2}R$.
     
     
     • For diatomic gas: $C_v = frac{5}{2}R$.
     
     
     
     
   
   
   • Work Done (W = Work done ON the system):
     
     
     
     • General: $W = int P dV$. (Area under P-V curve, with appropriate sign based on direction)
     
     
     • Isobaric: $W = P(V_2 - V_1)$.
     
     
     • Isochoric: $W = 0$.
     
     
     • Isothermal: $W = nRT ln(V_2/V_1) = nRT ln(P_1/P_2)$.
     
     
     • Adiabatic: $W = frac{P_2 V_2 - P_1 V_1}{gamma - 1} = frac{nR(T_2 - T_1)}{gamma - 1}$.
     
     
     
     
   
   
   • Heat (Q):
     
     
     
     • General: $Q = n C Delta T$ (where $C$ is molar specific heat).
     
     
     • Isochoric: $Q = n C_v Delta T = Delta U$.
     
     
     • Isobaric: $Q = n C_p Delta T$.
     
     
     • Adiabatic: $Q = 0$.
     
     
     • Isothermal: $Q = -W$ (since $Delta U = 0$).
     
     
     
     
   
   
   
   


5. 
   Substitute and Solve:
   
   
   
   • Plug in the given values and the derived expressions into the First Law equation $Delta U = Q + W$.
   
   
   • Solve for the unknown quantity. Ensure all units are consistent (e.g., Joules for energy, Pascals for pressure, m³ for volume).
   
   
   
   

Common Pitfalls (JEE & CBSE):

• Sign Errors: This is by far the most common mistake. Carefully apply sign conventions for $Q$ and $W$.


• Mixing Conventions: Do not mix conventions for work (i.e., work done *by* the system vs. *on* the system) within the same problem. Stick to one.


• Incorrect Process Identification: Assuming a process is isothermal when it's not, or misapplying adiabatic relations.


• Internal Energy for Non-Ideal Gases: For ideal gases, $Delta U$ only depends on $Delta T$. This is not true for non-ideal gases or phase changes. (JEE mostly deals with ideal gases).


• Units: Always convert all quantities to SI units (Joules, Pascals, m³, Kelvin) before calculations. $R$ value must also be in J/mol·K.

By following these steps meticulously, you can confidently approach and solve problems based on the First Law of Thermodynamics.

For CBSE board examinations, the First Law of Thermodynamics is a foundational concept. The focus is primarily on its definition, mathematical representation, crucial sign conventions, and direct application to different thermodynamic processes. Understanding these aspects clearly is vital for scoring well.

1. The First Law of Thermodynamics: Statement and Mathematical Form

• Statement: The First Law of Thermodynamics is essentially the law of conservation of energy applied to thermodynamic systems. It states that energy can neither be created nor destroyed; it can only be transformed from one form to another.


• Mathematical Form:
  
  
  
  • $Delta U = Q - W$
  
  
  • Where:
    
    
    
    • $Delta U$ = Change in internal energy of the system
    
    
    • $Q$ = Heat supplied to or rejected by the system
    
    
    • $W$ = Work done by or on the system
    
    
    
    
  
  
  • CBSE Note: This is the most commonly used form. Some texts might use $Delta U = Q + W$, where W represents work done *on* the system. Always be consistent with your chosen sign convention.
  
  
  
  

2. Crucial Sign Conventions (Most Important for CBSE)

Incorrect sign conventions are a common source of error in CBSE numerical problems. Master these:

• For Heat (Q):
  
  
  
  • Positive (+Q): Heat absorbed by the system (system gains energy).
  
  
  • Negative (-Q): Heat rejected by the system (system loses energy).
  
  
  
  


• For Work (W):
  
  
  
  • Positive (+W): Work done by the system (e.g., expansion of gas). This means the system loses energy by doing work on surroundings.
  
  
  • Negative (-W): Work done on the system (e.g., compression of gas). This means the surroundings do work on the system, increasing its energy.
  
  
  
  

3. Application to Specific Thermodynamic Processes

CBSE frequently asks for the application of the first law to ideal gas processes. Be prepared to state the first law for each:

• Isothermal Process ($T = ext{constant}$):
  
  
  
  • Since temperature is constant, and for an ideal gas internal energy $U$ depends only on $T$, then $Delta U = 0$.
  
  
  • Applying First Law: $0 = Q - W implies mathbf{Q = W}$.
  
  
  • Meaning: All heat absorbed by the system is converted into work done by the system.
  
  
  
  


• Adiabatic Process ($Q = 0$):
  
  
  
  • No heat exchange with the surroundings.
  
  
  • Applying First Law: $Delta U = 0 - W implies mathbf{Delta U = -W}$.
  
  
  • Meaning: Work is done at the expense of internal energy (expansion) or internal energy increases due to work done on the system (compression).
  
  
  
  


• Isochoric Process ($V = ext{constant}$):
  
  
  
  • Since volume is constant, no work is done (W = P$Delta V$ = 0).
  
  
  • Applying First Law: $Delta U = Q - 0 implies mathbf{Delta U = Q}$.
  
  
  • Meaning: All heat supplied to the system increases its internal energy.
  
  
  
  


• Isobaric Process ($P = ext{constant}$):
  
  
  
  • Work done W = P$Delta V$.
  
  
  • Applying First Law: $mathbf{Delta U = Q - PDelta V}$.
  
  
  • Meaning: Heat supplied changes both the internal energy and does work.
  
  
  
  


• Cyclic Process:
  
  
  
  • The system returns to its initial state, so the net change in internal energy is zero ($Delta U_{cycle} = 0$).
  
  
  • Applying First Law: $0 = Q_{net} - W_{net} implies mathbf{Q_{net} = W_{net}}$.
  
  
  • Meaning: The net heat exchanged in a cycle equals the net work done.
  
  
  
  

4. Key Takeaways for CBSE Exams

• Definitions: Be able to define Internal Energy, Heat, and Work. Understand that $U$ is a state function, while $Q$ and $W$ are path functions.


• Derivations: Derivations for work done in isothermal and adiabatic processes for an ideal gas are frequently asked (e.g., $W = nRT ln(V_f/V_i)$ for isothermal).


• Numerical Problems: Expect straightforward numerical problems where you are given values for two of the three variables ($Delta U, Q, W$) and asked to calculate the third, often for a specific process or a simple cycle, ensuring correct sign conventions are applied.

Pro Tip: Practice drawing P-V diagrams for each process. Visualizing the work done (area under the curve) helps immensely in understanding the first law applications.