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Step-by-Step Solution
Step 1: Identify the Processes
The gas undergoes two distinct processes one after the other:
Isothermal Expansion from volume $V$ to $2V$.
Adiabatic Expansion from volume $2V$ to $16V$.
Given quantities:
Initial pressure: $P$
Initial volume: $V$
Ratio of specific heats ($\gamma$): $\frac{5}{3}$
Step 2: Apply the Isothermal Process Relation
For an isothermal process (constant temperature), the product of pressure and volume remains constant, i.e. $PV = \text{constant}$. Hence,
$P \times V = P' \times (2V)$
Solving for the intermediate pressure $P'$ (the pressure at the final state of the isothermal process):
$P' = \frac{P \times V}{2V} = \frac{P}{2}
Step 3: Use the Adiabatic Condition
For an adiabatic expansion (no heat exchange), the quantity $P V^\gamma$ remains constant. Thus, if $P_f$ is the final pressure after the adiabatic process, we have:
$P' \,(2V)^\gamma = P_f \,(16V)^\gamma
Substitute $P' = \frac{P}{2}$ and simplify:
$\frac{P}{2} \times (2V)^\gamma = P_f \times (16V)^\gamma
Step 4: Express Volumes in the Adiabatic Relation
We note that $(2V)^\gamma = 2^\gamma \, V^\gamma$ and $(16V)^\gamma = 16^\gamma \, V^\gamma.$
So,
$\frac{P}{2} \times 2^\gamma \, V^\gamma = P_f \times 16^\gamma \, V^\gamma
The $V^\gamma$ terms cancel out on both sides, giving:
$\frac{P}{2} \times 2^\gamma = P_f \times 16^\gamma
But $16 = 2^4$, so $16^\gamma = (2^4)^\gamma = 2^{4\gamma}.$ Hence the equation becomes:
$\frac{P}{2} \times 2^\gamma = P_f \times 2^{4\gamma}
Step 5: Solve for the Final Pressure $P_f$
$P_f = \frac{\frac{P}{2} \times 2^\gamma}{2^{4\gamma}} = \frac{P}{2} \times 2^{\gamma - 4\gamma} = \frac{P}{2} \times 2^{-3\gamma}
Since $\gamma = \frac{5}{3}, \; -3\gamma = -3 \times \frac{5}{3} = -5.$ Thus,
$P_f = \frac{P}{2} \times 2^{-5} = \frac{P}{2} \times \frac{1}{32} = \frac{P}{64}.
Final Answer
The final pressure of the gas is $\frac{P}{64}$.
