Question

When an ideal gas $(\gamma = \frac{5}{3})$ is heated under constant pressure, then what percentage of given heat energy will be utilised in doing external work?

• A. $60%$
• B. $20%$
• C. $30%$
• D. $40%$

Hint:

For heating at constant pressure: \[ \frac{W}{Q}=\frac{R}{C_P} \] So first find \(C_P\) from \(\gamma\), then directly calculate the fraction.

Answer 1

The Correct Option is D

Concept:
At constant pressure: \[ Q=nC_P\Delta T \] and work done is: \[ W=P\Delta V=nR\Delta T \] So the fraction of heat used in external work is: \[ \frac{W}{Q}=\frac{nR\Delta T}{nC_P\Delta T}=\frac{R}{C_P} \] ip

Step 1: Relate \(C_P\) with \(\gamma\).
We know: \[ \gamma=\frac{C_P}{C_V}=\frac{5}{3} \] and \[ C_P-C_V=R \] Also, \[ C_P=\frac{\gamma R}{\gamma-1} \] Substitute \(\gamma=\frac53\): \[ C_P=\frac{\frac53 R}{\frac53-1} =\frac{\frac53 R}{\frac23} =\frac52 R \] ip

Step 2: Find the fraction of heat used as work.
\[ \frac{W}{Q}=\frac{R}{C_P} =\frac{R}{\frac52 R} =\frac{2}{5} \] \[ \frac{W}{Q}=0.4 \] ip

Step 3: Convert into percentage.
\[ 0.4\times 100 = 40% \] ip Hence, the correct answer is:
\[ \boxed{(D)\ 40%} \]