Analysis
The system is perfectly insulated (Adiabatic, $Q=0$). The external pressure suddenly triples ($P_{ext} = 3P_0$), compressing the gas until the piston stops. This is an irreversible adiabatic process.
First Law of Thermodynamics: $\Delta U = W_{on\_gas}$
The work done on the gas by the constant external pressure is:
$$ W = P_{ext} (V_i – V_f) = 3P_0 (V_0 – V) $$The change in internal energy for an ideal gas is:
$$ \Delta U = \frac{P_f V_f – P_i V_i}{\gamma – 1} $$At equilibrium, the final pressure of the gas equals the external pressure: $P_f = 3P_0$.
$$ \Delta U = \frac{3P_0 V – P_0 V_0}{\gamma – 1} $$Calculation
Equating $\Delta U$ and $W$:
$$ \frac{P_0 (3V – V_0)}{\gamma – 1} = 3P_0 (V_0 – V) $$For Helium (mono-atomic), $\gamma = 5/3$, so $\gamma – 1 = 2/3$.
$$ \frac{3V – V_0}{2/3} = 3(V_0 – V) $$ $$ \frac{3}{2} (3V – V_0) = 3(V_0 – V) $$Divide by 3:
$$ \frac{3V – V_0}{2} = V_0 – V $$ $$ 3V – V_0 = 2V_0 – 2V $$ $$ 5V = 3V_0 \implies V = \frac{3}{5} V_0 $$Correct Option: (b) 3/5