Question

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

• A. 64P
• B. 32P
• C. \(\frac{P}{64}\)
• D. 16P

Answer 1

The Correct Option is C

 To solve this problem, we need to understand the thermodynamic processes involved: an isothermal expansion followed by an adiabatic expansion.

Given: A monoatomic gas with an initial pressure \(P\) and volume \(V\). It expands isothermally to a volume \(2V\) and then adiabatically to \(16V\). The adiabatic index \(\gamma = \frac{5}{3}\).

Step 1: Isothermal Expansion

In an isothermal process, the temperature of the system remains constant. For an ideal gas undergoing an isothermal process, the relationship between the initial and final states is given by Boyle's Law:

\(P_1 V_1 = P_2 V_2\), where \(P_1 = P\), \(V_1 = V\), \(V_2 = 2V\)

So, substituting the given values:

\(P \cdot V = P_2 \cdot 2V \implies P_2 = \frac{P}{2}\)

Step 2: Adiabatic Expansion

In an adiabatic process, no heat is exchanged with the surroundings. For an ideal gas, the pressure and volume are related as:

\(P_2 V_2^{\gamma} = P_3 V_3^{\gamma}\)

Given that \(V_2 = 2V\), \(V_3 = 16V\), and \(\gamma = \frac{5}{3}\), and substituting \(P_2 = \frac{P}{2}\), we have:

\(\left(\frac{P}{2}\right) \cdot (2V)^{\frac{5}{3}} = P_3 \cdot (16V)^{\frac{5}{3}}\)

To find \(P_3\), we simplify the equation:

\(\frac{P}{2} \cdot 2^{\frac{5}{3}}V^{\frac{5}{3}} = P_3 \cdot (16V)^{\frac{5}{3}}\)

Simplifying further, note that:

\(2^{\frac{5}{3}} = 4^{\frac{5}{3} \over 2} \quad \text{and} \quad 16V = (2^4)V\)

After simplification:

\(2^{\frac{5}{3}} \times \frac{P}{2} = P_3 \times (2^4)^{\frac{5}{3}}\)

(2^{\frac{5}{3}} \cdot \frac{P}{2}) = P_3 \cdot 2^{\frac{20}{3}}

So the final pressure is:

\(P_3 = \frac{P}{64}\)

Conclusion: The correct answer is \(\frac{P}{64}\).