Question

A monoatomic gas at pressure ‘P’ having volume ‘V’ expands isothermally to a volume 2 V and then adiabatically to a volume 16 V. The final pressure of the gas is \(\gamma = \frac{5}{3}\)

• A. \(\frac{P}{64}\)
• B. \(\frac{P}{128}\)
• C. \(\frac{P}{8}\)
• D. \(\frac{P}{32}\)

Hint:

In problems with multiple thermodynamic processes, handle each step sequentially. For adiabatic processes, remember \(TV^{\gamma-1} = \text{constant}\) or \(PV^{\gamma} = \text{constant}\). Here using \(PV^{\gamma}\) is direct.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
A monoatomic ideal gas (\(\gamma = 5/3\)) first expands isothermally from volume \(V\) to \(2V\), then expands adiabatically from \(2V\) to \(16V\). We need the final pressure in terms of the initial pressure \(P\).

Step 2: Key Formula or Approach:
For an isothermal process: \(P V = \text{constant}\).
For an adiabatic process: \(P V^{\gamma} = \text{constant}\).

Step 3: Detailed Explanation:
Isothermal expansion:
\(P_1 = P\), \(V_1 = V\), \(V_2 = 2V\).
\(P_1 V_1 = P_2 V_2 \quad \Rightarrow \quad P \cdot V = P_2 \cdot (2V) \quad \Rightarrow \quad P_2 = \frac{P}{2}\).
Adiabatic expansion:
Initial for adiabatic step: \(P_2 = \frac{P}{2}\), \(V_2 = 2V\). Final volume \(V_3 = 16V\).
\(P_2 V_2^{\gamma} = P_3 V_3^{\gamma}\).
\[ P_3 = P_2 \left( \frac{V_2}{V_3} \right)^{\gamma} = \frac{P}{2} \left( \frac{2V}{16V} \right)^{5/3} = \frac{P}{2} \left( \frac{1}{8} \right)^{5/3}. \] Now \(\left( \frac{1}{8} \right)^{5/3} = 8^{-5/3} = (2^3)^{-5/3} = 2^{-5} = \frac{1}{32}\).
Thus \(P_3 = \frac{P}{2} \times \frac{1}{32} = \frac{P}{64}\).

Step 4: Final Answer:
The final pressure is \(\frac{P}{64}\), option (A).