Question

A monoatomic gas at pressure P and volume V is suddenly compressed to one eighth of its original volume. The final pressure at constant entropy will be :

• A. P
• B. 8P
• C. 32P
• D. 64P

Answer 1

The Correct Option is C

To solve this problem, we need to understand the behavior of a monoatomic gas under adiabatic processes. When a gas is compressed or expanded without any heat exchange with the environment, it undergoes an adiabatic process. For a monoatomic ideal gas, this process follows the equation:

PV^{\gamma} = \text{constant}

where \gamma, the adiabatic index, is the ratio of specific heats and equals \frac{5}{3} for a monoatomic gas.

1. Initially, the gas is at pressure P and volume V.
2. The volume is reduced to one eighth of its original value, so the final volume V_f = \frac{V}{8}.
3. Applying the adiabatic relation:

P_i V_i^{\gamma} = P_f V_f^{\gamma}

Substituting the known values:

P \cdot V^{\frac{5}{3}} = P_f \cdot \left(\frac{V}{8}\right)^{\frac{5}{3}}

Solving for the final pressure P_f:

P_f = P \cdot \left(\frac{V}{\frac{V}{8}}\right)^{\frac{5}{3}} = P \cdot (8)^{\frac{5}{3}}

Calculating (8)^{\frac{5}{3}}:

• (8)^{\frac{5}{3}} = \left(2^3\right)^{\frac{5}{3}} = 2^5 = 32

Therefore, the final pressure P_f = 32P.

Thus, the correct answer is 32P.