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

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: $\gamma = \frac{5}{3}$)

Accepted Answer

The process consists of two steps:

**Step 1: Isothermal Expansion**
Initial state: $P_1 = P, V_1 = V$
Final state of step 1: $V_2 = 2V$
For an isothermal process, $PV = 	ext{constant}$ (Boyle's Law).
$P_1 V_1 = P_2 V_2$
$P \cdot V = P_2 \cdot (2V)$
$P_2 = \frac{P}{2}$

**Step 2: Adiabatic Expansion**
Initial state of step 2: $P_2 = \frac{P}{2}, V_2 = 2V$
Final state: $V_3 = 16V$
For an adiabatic process, $PV^{\gamma} = 	ext{constant}$.
$P_2 V_2^{\gamma} = P_3 V_3^{\gamma}$
Substituting the values:
$(\frac{P}{2}) (2V)^{\gamma} = P_3 (16V)^{\gamma}$
$P_3 = \frac{P}{2} \left( \frac{2V}{16V} ight)^{\gamma}$
$P_3 = \frac{P}{2} \left( \frac{1}{8} ight)^{\frac{5}{3}}$
$P_3 = \frac{P}{2} \left( (\frac{1}{2})^3 ight)^{\frac{5}{3}}$
$P_3 = \frac{P}{2} \left( \frac{1}{2} ight)^{5}$
$P_3 = \frac{P}{2} \cdot \frac{1}{32} = \frac{P}{64}$

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