Question

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

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

Answer 1

Answer: (A)

$40$%

For isobaric process:
$\Delta Q=\Delta U+\Delta W$
But, $\Delta Q=n\left(\frac{\gamma }{\gamma -1}\right)R\Delta T$
$\Delta U=n\frac{R}{\gamma -1}\Delta T$
$\therefore \frac{\Delta U}{\Delta Q}=\frac{\left(\frac{1}{\gamma -1}\right)}{\left(\frac{\gamma }{\gamma -1}\right)}=\frac{1}{\gamma }=\frac{3}{5}$ as $\gamma =\frac{5}{3}$
Hence, $\frac{\Delta W}{\Delta Q}=1-\frac{\Delta U}{\Delta Q}=1-\frac{3}{5}=\frac{2}{5}$
Thus $\frac{2}{5}\times 100$%$=40$% of the heat absorbed is used in doing external work.