Solution for Question 31
Physics Principle:
The problem states the gas expands quasi-statically due to friction. Since the walls are conducting and the process is slow (quasi-static), the expansion is Isothermal at temperature $T$.
Energy Balance:
According to the Work-Energy Theorem for the massless piston (kinetic energy change is zero):
$$W_{gas} + W_{atm} + W_{friction} = 0$$
1. Work done by Gas ($W_{gas}$):
For a reversible isothermal expansion from $P_1$ to $p_0$:
$$W_{gas} = \int P \, dV = NkT \ln \left( \frac{V_f}{V_i} \right) = NkT \ln \left( \frac{P_1}{p_0} \right)$$
2. Work done by Atmosphere ($W_{atm}$):
The atmosphere exerts a constant pressure $p_0$ opposing the expansion.
$$W_{atm} = -p_0 (V_f – V_i)$$
Using $V = NkT/P$:
$$W_{atm} = -p_0 \left( \frac{NkT}{p_0} – \frac{NkT}{P_1} \right) = -NkT \left( 1 – \frac{p_0}{P_1} \right)$$
3. Work done by Friction ($W_{friction}$):
From the energy balance equation:
$$W_{friction} = -(W_{gas} + W_{atm})$$
$$W_{friction} = -\left[ NkT \ln \left( \frac{P_1}{p_0} \right) – NkT \left( 1 – \frac{p_0}{P_1} \right) \right]$$
$$W_{friction} = NkT \left[ \left( 1 – \frac{p_0}{P_1} \right) – \ln \left( \frac{P_1}{p_0} \right) \right]$$
$$W_{friction} = NkT \left[ 1 – \frac{p_0}{P_1} – \ln \left( \frac{P_1}{p_0} \right) \right]$$