Question

When a polyatomic gas is heated at constant pressure, the percentage of heat given to the gas that is converted into external work is
(Ratio of the specific heat capacities of the gas = $\frac{4}{3}$)

• A. 30
• B. 25
• C. 20
• D. 45

Hint:

For a gas at constant pressure, the fraction of heat converted to work depends on the specific heat ratio $\gamma$, given by $\frac{\gamma - 1}{\gamma}$.

Answer 1

The Correct Option is B

For a gas heated at constant pressure, the heat supplied \( Q = nC_p \Delta T \), where \( C_p \) is the molar specific heat at constant pressure.

The work done is \( W = P \Delta V = nR \Delta T \) (using the ideal gas law).

The fraction of heat converted into work is \( \frac{W}{Q} = \frac{nR \Delta T}{nC_p \Delta T} = \frac{R}{C_p} \).

The ratio of specific heats is \( \gamma = \frac{C_p}{C_v} = \frac{4}{3} \). Since \( C_p - C_v = R \), we have \( C_p = \gamma C_v \) and \( C_v = \frac{C_p}{\gamma} \). Thus,
\( C_p - \frac{C_p}{\gamma} = R \implies C_p \left(1 - \frac{1}{\gamma}\right) = R \implies C_p \left(\frac{\gamma - 1}{\gamma}\right) = R \implies \frac{R}{C_p} = \frac{\gamma - 1}{\gamma} \).

Substituting \( \gamma = \frac{4}{3} \):
\( \frac{R}{C_p} = \frac{\frac{4}{3} - 1}{\frac{4}{3}} = \frac{\frac{1}{3}}{\frac{4}{3}} = \frac{1}{4} = 0.25 \).
The percentage is \( 0.25 \times 100 = 25\% \).