Question

A monoatomic gas at pressure P having volume V expands isothermally to a volume 3V and then adiabatically to a volume 81 V. The final pressure of the gas is (\(γ = 5/3\)).

• A. \(\frac{\text{P}}{3}\)
• B. \(\frac{\text{P}}{(3)^2}\)
• C. \(\frac{\text{P}}{(3)^4}\)
• D. \(\frac{\text{P}}{(3)^6}\)

Hint:

Use Boyle's law for the isothermal step and PV^gamma for the adiabatic step.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Isothermal: \(PV = \text{constant}\). Adiabatic: \(PV^\gamma = \text{constant}\) with \(\gamma = \tfrac53\).

Step 2: Isothermal step:
\(P\cdot V = P_1\cdot3V \Rightarrow P_1 = \dfrac P3\).

Step 3: Adiabatic step:
\[ P_1(3V)^{5/3} = P_2(81V)^{5/3} \Rightarrow P_2 = P_1\left(\frac{3}{81}\right)^{5/3} = P_1\left(\frac1{27}\right)^{5/3} \]
\(27^{5/3} = (3^3)^{5/3} = 3^5\), so \(P_2 = \dfrac{P_1}{3^5}\).

Step 4: Result:
\[ P_2 = \frac{P}{3}\cdot\frac1{3^5} = \frac{P}{3^6} \]
Option (D). Options (A), (B), (C) have too small a power of 3.

Final Answer:
P1 = P/3, then dividing by 3^5 gives P/3^6. \[ \boxed{\text{(D) }\dfrac{P}{3^6}} \]