First Law of Thermodynamics

Overview

The first law of thermodynamics is a statement of energy conservation for thermal systems. It relates heat, work, and internal energy.

Statement

$$\Delta U = Q - W$$

Or equivalently:

$$Q = \Delta U + W$$

Where:

• $\Delta U$ = change in internal energy
• $Q$ = heat added to the system
• $W$ = work done BY the system

Sign Conventions

Quantity	Positive	Negative
$Q$	Heat absorbed	Heat released
$W$	Work done by system	Work done on system
$\Delta U$	Energy increases	Energy decreases

Internal Energy

• Total kinetic and potential energy of molecules
• Depends only on temperature (for ideal gas)
• State function (path independent)

For ideal monatomic gas:

$$U = \frac{3}{2}nRT$$

For ideal diatomic gas:

$$U = \frac{5}{2}nRT$$

Change in internal energy:

$$\Delta U = nC_v\Delta T$$

Work Done by a Gas

General Expression

$$W = \int P \, dV$$

Work equals area under P-V curve

Constant Pressure (Isobaric)

$$W = P\Delta V = P(V_2 - V_1)$$

Constant Volume (Isochoric)

$$W = 0$$

Isothermal (Constant Temperature)

$$W = nRT \ln\left(\frac{V_2}{V_1}\right) = nRT \ln\left(\frac{P_1}{P_2}\right)$$

Adiabatic (No Heat Transfer)

$$W = \frac{P_1V_1 - P_2V_2}{\gamma - 1} = nC_v(T_1 - T_2)$$

Thermodynamic Processes

Isothermal Process ($\Delta T = 0$)

• Temperature constant
• $\Delta U = 0$
• $Q = W$
• $PV = \text{constant}$

Isobaric Process ($\Delta P = 0$)

• Pressure constant
• $W = P\Delta V$
• $Q = nC_p\Delta T$

Isochoric Process ($\Delta V = 0$)

• Volume constant
• $W = 0$
• $Q = \Delta U = nC_v\Delta T$

Adiabatic Process ($Q = 0$)

• No heat exchange
• $\Delta U = -W$
• $PV^\gamma = \text{constant}$
• $TV^{\gamma-1} = \text{constant}$

Heat Capacities of Gases

At Constant Volume

$$C_v = \frac{f}{2}R$$

At Constant Pressure

$$C_p = C_v + R = \frac{f+2}{2}R$$

Where $f$ = degrees of freedom:

• Monatomic: $f = 3$, $C_v = \frac{3}{2}R$, $C_p = \frac{5}{2}R$
• Diatomic: $f = 5$, $C_v = \frac{5}{2}R$, $C_p = \frac{7}{2}R$

Ratio of Heat Capacities

$$\gamma = \frac{C_p}{C_v} = \frac{f+2}{f}$$

• Monatomic: $\gamma = \frac{5}{3} \approx 1.67$
• Diatomic: $\gamma = \frac{7}{5} = 1.4$

Adiabatic Relations

$$PV^\gamma = \text{constant}$$ $$TV^{\gamma-1} = \text{constant}$$ $$T^\gamma P^{1-\gamma} = \text{constant}$$

Examples

Example 1: Isochoric Heating

2 moles of monatomic gas at 300 K is heated at constant volume to 500 K.

$$\Delta U = nC_v\Delta T = 2 \times \frac{3}{2} \times 8.314 \times 200 = 4988 \text{ J}$$ $$W = 0 \quad \text{(constant volume)}$$ $$Q = \Delta U = 4988 \text{ J}$$

Example 2: Isobaric Expansion

2 moles of diatomic gas expands at 1 atm from 300 K to 400 K.

$$\Delta V = \frac{nR\Delta T}{P} = \frac{2 \times 8.314 \times 100}{101325} = 0.0164 \text{ m}^3$$ $$W = P\Delta V = 101325 \times 0.0164 = 1663 \text{ J}$$ $$\Delta U = nC_v\Delta T = 2 \times \frac{5}{2} \times 8.314 \times 100 = 4157 \text{ J}$$ $$Q = \Delta U + W = 5820 \text{ J}$$

Example 3: Isothermal Expansion

1 mole of gas at 300 K expands from 1 L to 3 L isothermally.

$$W = nRT \ln\left(\frac{V_2}{V_1}\right) = 1 \times 8.314 \times 300 \times \ln(3) = 2740 \text{ J}$$ $$\Delta U = 0 \quad \text{(isothermal)}$$ $$Q = W = 2740 \text{ J}$$

Example 4: Adiabatic Compression

Monatomic gas ($\gamma = \frac{5}{3}$) at 300 K, 1 atm is compressed to 1/8 original volume.

$$T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma-1} = 300 \times 8^{2/3} = 300 \times 4 = 1200 \text{ K}$$ $$P_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma = 1 \times 8^{5/3} = 32 \text{ atm}$$

Example 5: Cyclic Process

A gas undergoes a cyclic process. Total heat absorbed is 1000 J.

For a complete cycle:

$$\Delta U = 0 \quad \text{(returns to initial state)}$$ $$Q = W = 1000 \text{ J} \quad \text{(net work done)}$$

Free Expansion

Gas expands into vacuum:

• $W = 0$ (no external pressure)
• $Q = 0$ (too fast for heat transfer)
• $\Delta U = 0$
• For ideal gas: $\Delta T = 0$