First Law of Thermodynamics

The First Law of Thermodynamics is essentially a statement of the conservation of energy, applied to thermodynamic systems. It states that:

\( \Delta U = Q - W \)

where:

\( \Delta U \): Change in internal energy of the system
\( Q \): Heat added to the system
\( W \): Work done by the system on its surroundings

Be cautious with signs:

Heat added to the system is positive
Work done by the system is positive

The internal energy \( U \) of an ideal gas depends only on its temperature. For a monatomic ideal gas:
\( U = \frac{3}{2}nRT \)

Thermodynamic Processes

Depending on how heat, work, and internal energy interact, we classify processes as follows:

1. Isochoric Process (Constant Volume)

Volume does not change → \( \Delta V = 0 \), so \( W = 0 \)
First Law simplifies to \( \Delta U = Q \)
All the heat added changes internal energy (temperature)

2. Isobaric Process (Constant Pressure)

Pressure is constant → work is done as volume changes
\( W = P \Delta V \)
First Law: \( Q = \Delta U + W \)
Heat changes both internal energy and does work

3. Isothermal Process (Constant Temperature)

Temperature constant → \( \Delta U = 0 \)
First Law: \( Q = W \)
Heat added is entirely used to do work

4. Adiabatic Process (No Heat Exchange)

\( Q = 0 \)
First Law: \( \Delta U = -W \)
If the system does work, it loses internal energy

2 moles of an ideal monatomic gas are heated at constant volume from \( 300 \, \text{K} \) to \( 600 \, \text{K} \). Find:

(a) Heat added to the gas
(b) Work done
(c) Change in internal energy

Take \( R = 8.31 \, \text{J/mol·K} \).

Solution:

Since volume is constant, \( W = 0 \)
\( \Delta U = \frac{3}{2}nR\Delta T = \frac{3}{2} \cdot 2 \cdot 8.31 \cdot (600 - 300) \)
\( \Delta U = 3 \cdot 8.31 \cdot 300 = 7479 \, \text{J} \)

(a) \( Q = \Delta U = 7479 \, \text{J} \)
(b) \( W = 0 \)
(c) \( \Delta U = 7479 \, \text{J} \)

A gas is compressed adiabatically, doing 500 J of work on it. During the compression, 200 J of heat is lost to the environment. What is the change in internal energy?

Solution:

Heat lost by system: \( Q = -200 \, \text{J} \)
Work done on the gas: \( W = -500 \, \text{J} \)
\( \Delta U = Q - W = -200 - (-500) = 300 \, \text{J} \)

An ideal gas is heated at constant volume with 1200 J of heat. What is the change in internal energy and the work done?

Solution:

Constant volume → \( W = 0 \)
So: \( \Delta U = Q = 1200 \, \text{J} \)

Work done: \( 0 \), Internal energy increase: \( 1200 \, \text{J} \)

Written by Thenura Dilruk