A monoatomic gas at a pressure $p$, having a volume $V$ expands isothermally to a volume $2V$ and then adiabat

Question

A monoatomic gas at a pressure $p$, having a volume $V$ expands isothermally to a volume $2V$ and then adiabatically to a volume $16V$. The final pressure of the gas is: (take $\gamma = 5/3$)

Accepted Answer

The problem involves two stages of expansion:

**Stage 1: Isothermal Expansion**
Process: $PV = 	ext{constant}$ (since $T$ is constant).
Initial State: $P_1 = p, V_1 = V$.
Final State: $V_2 = 2V$.
Using Boyle's Law: $P_1V_1 = P_2V_2$
$p \cdot V = P_2 \cdot 2V \implies P_2 = \frac{p}{2}$.

**Stage 2: Adiabatic Expansion**
Process: $PV^\gamma = 	ext{constant}$ (where $\gamma = 5/3$ for monoatomic gas).
Initial State: $P_2 = \frac{p}{2}, V_2 = 2V$.
Final State: $V_3 = 16V$.
Using the adiabatic relation: $P_2V_2^\gamma = P_3V_3^\gamma$
$\frac{p}{2} (2V)^{5/3} = P_3 (16V)^{5/3}$
$P_3 = \frac{p}{2} \left( \frac{2V}{16V} ight)^{5/3} = \frac{p}{2} \left( \frac{1}{8} ight)^{5/3}$
Calculate $(1/8)^{5/3}$: Since $8 = 2^3$, $(1/2^3)^{5/3} = 1/2^5 = 1/32$.
$P_3 = \frac{p}{2} \cdot \frac{1}{32} = \frac{p}{64}$.

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