THERMAL O30

Solution for Question 30

Process Analysis:

1. Initial State: Pressure $P_1$, Temperature $T$.
2. Expansion: The weight is removed. The gas expands rapidly against the constant external atmospheric pressure $p_0$.
3. Final State: The walls are conducting, and after settling (“finally”), the gas returns to thermal equilibrium with the surroundings at temperature $T$. The final pressure balances atmospheric pressure ($p_0$).

Work Done Calculation:

For expansion against a constant external pressure $p_0$, the work done by the gas is:

$$W = p_0 \Delta V = p_0 (V_{final} – V_{initial})$$

Since the initial and final temperatures are both $T$, we can use the Ideal Gas Law ($PV = NkT$):

• $V_{final} = \frac{NkT}{p_0}$
• $V_{initial} = \frac{NkT}{P_1}$

Substitute these into the work equation:

$$W = p_0 \left( \frac{NkT}{p_0} – \frac{NkT}{P_1} \right)$$

$$W = NkT – p_0 \frac{NkT}{P_1}$$

$$W = NkT \left( 1 – \frac{p_0}{P_1} \right)$$