First Law of Thermodynamics

Real-Life Hook: Why Do Bike Pumps Heat Up?

Ever pumped air into a bicycle tire and noticed the pump getting hot? You haven’t added any heat—you’ve just compressed air! This is the First Law of Thermodynamics in action: energy doesn’t vanish; it transforms.

When you compress air (doing work on it), that work energy converts to internal energy, raising the temperature. Your hand pump is a mini thermodynamics laboratory!

Similarly, when you release air from a spray can, it feels cold. The gas does work expanding, losing internal energy, and cooling down.

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The First Law of Thermodynamics

Statement

Energy can neither be created nor destroyed; it can only be transformed from one form to another.

In thermodynamics language:

The heat supplied to a system goes partly to increase its internal energy and partly to do work against external forces.

Mathematical Form

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Sign Conventions (CRITICAL for JEE!)

Standard Sign Convention (Most Common in JEE)

Quantity	Positive (+)	Negative (-)
Heat (Q)	Heat absorbed by system	Heat released by system
Work (W)	Work done BY system (expansion)	Work done ON system (compression)
Internal Energy (ΔU)	Temperature increases	Temperature decreases

Memory Trick: “System’s Perspective”

Think from the system’s viewpoint:

• Getting heat? Happy! → Positive Q
• Releasing heat? Sad! → Negative Q
• Doing work (expanding)? Tired! → Positive W
• Work done on it (compressed)? Energized! → Negative W

Common Sign Convention Mistakes

❌ Mistake 1: Forgetting that W is work BY system

• Compression: W is NEGATIVE (work done ON system)
• Expansion: W is POSITIVE (work done BY system)

❌ Mistake 2: Mixing up ΔU = Q - W and ΔU = Q + W

• Use: $\Delta U = Q - W$ (when W is work BY system)
• Some books use: $\Delta U = Q + W'$ (where W’ is work ON system)
• For JEE: Stick to $\Delta U = Q - W$ with W as work BY system

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Internal Energy (U)

What is Internal Energy?

Internal energy is the total energy contained within a system:

• Kinetic energy of molecules (translational, rotational, vibrational)
• Potential energy of molecular interactions

Key Properties

1. State Function: U depends only on current state (T, P, V), NOT on path

2. For Ideal Gas: $U$ depends ONLY on temperature

   $$U = nC_VT$$

   where $C_V$ is molar heat capacity at constant volume

3. Change in Internal Energy:

   $$\Delta U = nC_V\Delta T$$

   This works for ALL processes (isothermal, adiabatic, etc.)

4. At Constant Temperature: For ideal gas, $\Delta U = 0$ (isothermal process)

Monatomic vs Diatomic Gases

Gas Type	$C_V$	$C_P$	$\gamma = C_P/C_V$
Monatomic	$\frac{3}{2}R$	$\frac{5}{2}R$	$\frac{5}{3} ≈ 1.67$
Diatomic	$\frac{5}{2}R$	$\frac{7}{2}R$	$\frac{7}{5} = 1.4$

Memory Trick: “Mono 3-5, Di 5-7” (numerators for $C_V$ and $C_P$ with R/2)

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Heat (Q)

Definition

Heat is energy transferred between systems due to temperature difference.

Important Points

1. Heat is NOT a state function (depends on path)
2. Heat always flows from higher T to lower T
3. Heat transfer modes: Conduction, Convection, Radiation (not covered here)
4. Measured in Joules (J) or calories (cal): 1 cal = 4.186 J

Heat Capacity

More on this in Specific Heat Capacity →

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Work (W)

Work in Thermodynamics

When a gas expands or compresses, it does work:

Geometric Interpretation

Work = Area under P-V curve

This is crucial for JEE problems involving P-V diagrams!

Work in Different Processes

1. Isobaric (Constant P): $W = P\Delta V = nR\Delta T$
2. Isothermal (Constant T): $W = nRT \ln\frac{V_2}{V_1} = nRT \ln\frac{P_1}{P_2}$
3. Isochoric (Constant V): $W = 0$ (no volume change!)
4. Adiabatic: $W = \frac{nR(T_1 - T_2)}{\gamma - 1} = \frac{P_1V_1 - P_2V_2}{\gamma - 1}$

More details in Thermodynamic Processes →

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Applications of First Law

Application 1: Isothermal Process (ΔT = 0)

For ideal gas: $\Delta U = nC_V\Delta T = 0$

From First Law:

$$Q = \Delta U + W = 0 + W$$ $$\boxed{Q = W}$$

All heat supplied is converted to work (or vice versa).

Application 2: Adiabatic Process (Q = 0)

No heat exchange with surroundings.

From First Law:

$$0 = \Delta U + W$$ $$\boxed{\Delta U = -W}$$

Work done comes from internal energy (temperature decreases during expansion).

Application 3: Isochoric Process (ΔV = 0)

No volume change, so $W = 0$.

From First Law:

$$\boxed{Q = \Delta U}$$

All heat goes to internal energy (temperature change).

Application 4: Isobaric Process (P = constant)

$$Q = \Delta U + W$$ $$Q = nC_V\Delta T + P\Delta V$$ $$Q = nC_V\Delta T + nR\Delta T$$ $$\boxed{Q = nC_P\Delta T}$$

where $C_P = C_V + R$ (Mayer’s equation).

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Cyclic Process

In a cyclic process, system returns to initial state:

• $\Delta U = 0$ (state function!)

From First Law:

$$Q = W$$

Net heat absorbed = Net work done

This is the basis of heat engines!

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Free Expansion (Joule Expansion)

Gas expands into vacuum (no external pressure):

• $Q = 0$ (isolated)
• $W = 0$ (no opposing pressure)
• Therefore: $\Delta U = 0$

For ideal gas: $\Delta T = 0$ (temperature doesn’t change!)

Real gases: Temperature may change slightly due to intermolecular forces.

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Common Mistakes & Tricks

Mistake 1: Sign of Work in Compression

❌ Gas compressed from 2L to 1L, $W = P(1-2) = -P$ L ✅ Correct interpretation: W is negative (work done ON system)

When V decreases → W is negative

Mistake 2: Using ΔU for Non-Ideal Processes

For ideal gas: $\Delta U = nC_V\Delta T$ ALWAYS (regardless of process)

But $Q$ and $W$ depend on the path!

Mistake 3: Adiabatic vs Isothermal

Process	Q	ΔU	W	ΔT
Isothermal	$nRT\ln(V_2/V_1)$	0	$nRT\ln(V_2/V_1)$	0
Adiabatic	0	$nC_V\Delta T$	$-nC_V\Delta T$	≠ 0

Memory Trick: “QUIT”

Q = U + W

• Quantity of heat
• Up goes internal energy
• Work done by system

Trick for Cyclic Processes

“Round trip? No energy change!”

• Cyclic → $\Delta U = 0$ → $Q = W$

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Practice Problems

Level 1: JEE Main Basics

Q1. A system absorbs 500 J of heat and does 200 J of work. Find the change in internal energy.

Solution

Given:

• $Q = +500$ J (heat absorbed)
• $W = +200$ J (work done BY system)

First Law: $\Delta U = Q - W$

$$\Delta U = 500 - 200 = 300 \text{ J}$$

Internal energy increases by 300 J.

Q2. In an isothermal expansion, an ideal gas does 300 J of work. How much heat is absorbed?

Solution

Isothermal → $\Delta T = 0$ → $\Delta U = 0$

First Law: $Q = \Delta U + W = 0 + 300$

$$Q = 300 \text{ J}$$

Heat absorbed = Work done in isothermal process.

Q3. A gas is compressed at constant pressure from 4 L to 2 L by an external pressure of 2 atm. Find work done. (1 atm = $10^5$ Pa)

Solution

$$W = P\Delta V = P(V_2 - V_1)$$ $$W = 2 \times 10^5 \times (2 - 4) \times 10^{-3}$$ $$W = 2 \times 10^5 \times (-2) \times 10^{-3}$$ $$W = -400 \text{ J}$$

Negative sign: Work done ON the system (compression).

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Level 2: JEE Main/Advanced

Q4. One mole of an ideal gas at 300 K is expanded isothermally from 1 L to 10 L. Find (a) change in internal energy, (b) work done, (c) heat absorbed. (R = 8.314 J/mol·K)

Solution

(a) Change in internal energy: Isothermal → $\Delta U = nC_V\Delta T = 0$ (since $\Delta T = 0$)

(b) Work done:

$$W = nRT\ln\frac{V_2}{V_1}$$ $$W = 1 \times 8.314 \times 300 \times \ln\frac{10}{1}$$ $$W = 2494.2 \times \ln(10)$$ $$W = 2494.2 \times 2.303 = 5744 \text{ J}$$

(c) Heat absorbed: First Law: $Q = \Delta U + W = 0 + 5744$

$$Q = 5744 \text{ J} ≈ 5.74 \text{ kJ}$$

Alternative: In isothermal process, $Q = W$ directly.

Q5. Two moles of a monatomic ideal gas undergo a process where temperature changes from 300 K to 400 K at constant volume. Find (a) $\Delta U$, (b) W, (c) Q.

Solution

For monatomic gas: $C_V = \frac{3}{2}R = \frac{3}{2} \times 8.314 = 12.47$ J/mol·K

(a) Change in internal energy:

$$\Delta U = nC_V\Delta T = 2 \times 12.47 \times (400 - 300)$$ $$\Delta U = 2 \times 12.47 \times 100 = 2494 \text{ J}$$

(b) Work done: Constant volume → $\Delta V = 0$ → $W = 0$

(c) Heat absorbed:

$$Q = \Delta U + W = 2494 + 0 = 2494 \text{ J}$$

At constant volume, all heat increases internal energy.

Q6. A gas undergoes a cyclic process ABCA as shown. In process AB, 400 J of heat is absorbed. In process BC, 200 J of work is done by the gas. In process CA, 150 J of heat is rejected. Find: (a) Work done in AB (b) Heat absorbed in BC (c) Work done in CA

Solution

For cyclic process: $\Delta U_{cycle} = 0$

Therefore: $Q_{net} = W_{net}$

(a) Work in AB: Need to find $W_{net}$ first.

$$Q_{net} = Q_{AB} + Q_{BC} + Q_{CA} = 400 + Q_{BC} - 150$$

For individual processes, assume:

• AB: isobaric or isothermal (not specified, need more info)

Let’s use the cyclic property:

$$Q_{AB} + Q_{BC} + Q_{CA} = W_{AB} + W_{BC} + W_{CA}$$ $$400 + Q_{BC} - 150 = W_{AB} + 200 + W_{CA}$$

We need more information. Let me reconsider:

Standard approach:

For the cycle: $Q_{total} = W_{total}$

$$Q_{total} = 400 + Q_{BC} - 150 = W_{total}$$

If we assume BC is adiabatic (common in problems):

$$Q_{BC} = 0$$ $$\Delta U_{BC} = -W_{BC} = -200 \text{ J}$$

Then: $Q_{total} = 400 + 0 - 150 = 250$ J So: $W_{total} = 250$ J

$$W_{AB} + W_{BC} + W_{CA} = 250$$ $$W_{AB} + 200 + W_{CA} = 250$$ $$W_{AB} + W_{CA} = 50$$

Need individual process details. Problem incomplete without process types.

General Lesson: Always identify process type (isothermal, adiabatic, etc.) for each leg!

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Level 3: JEE Advanced

Q7. One mole of an ideal gas ($\gamma = 1.4$) at 300 K and 1 atm undergoes adiabatic expansion until pressure becomes 0.5 atm. Find: (a) Final temperature (b) Work done by gas (c) Change in internal energy

Solution

For adiabatic process: $T_1P_1^{(\gamma-1)/\gamma} = T_2P_2^{(\gamma-1)/\gamma}$

Or: $T_2 = T_1\left(\frac{P_2}{P_1}\right)^{(\gamma-1)/\gamma}$

(a) Final temperature:

$$T_2 = 300 \times \left(\frac{0.5}{1}\right)^{(1.4-1)/1.4}$$ $$T_2 = 300 \times (0.5)^{0.4/1.4}$$ $$T_2 = 300 \times (0.5)^{2/7}$$

Calculate: $(0.5)^{2/7} = 0.5^{0.286} ≈ 0.82$

$$T_2 ≈ 300 \times 0.82 = 246 \text{ K}$$

(b) Work done: For adiabatic process:

$$W = \frac{nR(T_1 - T_2)}{\gamma - 1}$$ $$W = \frac{1 \times 8.314 \times (300 - 246)}{1.4 - 1}$$ $$W = \frac{8.314 \times 54}{0.4}$$ $$W = \frac{449}{0.4} = 1122.5 \text{ J}$$

(c) Change in internal energy: For adiabatic: $Q = 0$

$$\Delta U = Q - W = 0 - 1122.5 = -1122.5 \text{ J}$$

Or directly: $\Delta U = nC_V\Delta T$

For $\gamma = 1.4$: $C_V = \frac{R}{\gamma - 1} = \frac{8.314}{0.4} = 20.785$ J/mol·K

$$\Delta U = 1 \times 20.785 \times (246 - 300) = 20.785 \times (-54) = -1122.4 \text{ J}$$

Internal energy decreases (gas cools during expansion).

Q8. A thermally insulated container has two compartments separated by a partition. One compartment has 1 mole of ideal gas at 300 K, the other has 2 moles at 400 K. The partition is removed. Find final temperature if: (a) Both gases are monatomic (b) First is monatomic, second is diatomic

Solution

System is isolated: $Q = 0$ (thermally insulated) No external work: $W = 0$

Therefore: $\Delta U_{total} = 0$

$$\Delta U_1 + \Delta U_2 = 0$$ $$n_1C_{V1}(T_f - T_1) + n_2C_{V2}(T_f - T_2) = 0$$

(a) Both monatomic: $C_V = \frac{3}{2}R$ for both

$$1 \times \frac{3}{2}R(T_f - 300) + 2 \times \frac{3}{2}R(T_f - 400) = 0$$ $$\frac{3R}{2}[(T_f - 300) + 2(T_f - 400)] = 0$$ $$T_f - 300 + 2T_f - 800 = 0$$ $$3T_f = 1100$$ $$T_f = \frac{1100}{3} ≈ 366.7 \text{ K}$$

(b) First monatomic, second diatomic:

• Gas 1: $C_{V1} = \frac{3}{2}R$
• Gas 2: $C_{V2} = \frac{5}{2}R$

$$1 \times \frac{3}{2}R(T_f - 300) + 2 \times \frac{5}{2}R(T_f - 400) = 0$$ $$\frac{R}{2}[3(T_f - 300) + 5 \times 2(T_f - 400)] = 0$$ $$3T_f - 900 + 10T_f - 4000 = 0$$ $$13T_f = 4900$$ $$T_f = \frac{4900}{13} ≈ 376.9 \text{ K}$$

Key Concept: Final temperature depends on heat capacities (weighted average).

Q9. An ideal gas undergoes a process where $P \propto \frac{1}{V}$ (rectangular hyperbola). Show that this is an isothermal process for an ideal gas.

Solution

Given: $P \propto \frac{1}{V}$

This means: $P = \frac{k}{V}$ for some constant k

Therefore: $PV = k$ (constant)

For ideal gas: $PV = nRT$

If $PV = k$ (constant), then:

$$nRT = k = \text{constant}$$

Since $n$ and $R$ are constants:

$$T = \frac{k}{nR} = \text{constant}$$

Therefore, the process is isothermal!

This is why isothermal process follows $P_1V_1 = P_2V_2$ and appears as a rectangular hyperbola on P-V diagram.

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Connection to Other Topics

→ Thermodynamic Processes

The First Law applied to specific processes (isothermal, adiabatic, etc.): Learn more →

→ PV Diagrams

Work as area under curve, cyclic processes: PV Diagrams →

→ Heat Engines

Cyclic application of First Law, efficiency: Heat Engines →

→ Kinetic Theory

Microscopic basis for internal energy:

$$U = nC_VT = n \times \frac{f}{2}RT$$

Kinetic Theory →

→ Chemistry: Thermodynamics

Enthalpy, $\Delta H = \Delta U + P\Delta V$: Chemistry Thermodynamics →

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Quick Revision Formula Sheet

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JEE Strategy Tips

1. Always check sign convention first thing
2. Identify the process (isothermal, adiabatic, etc.)
3. For ideal gas: $\Delta U = nC_V\Delta T$ in ALL processes
4. Cyclic process: Remember $\Delta U = 0$
5. Work = Area under P-V curve (visual approach)
6. Free expansion: $Q = W = \Delta U = 0$ for ideal gas

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Next Topic: Specific Heat Capacity →

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Last updated: February 2025