If heat energy $\Delta Q$ is supplied to an ideal diatomic gas, the increase in internal energy is $\Delta…
If heat energy $\Delta Q$ is supplied to an ideal diatomic gas, the increase in internal energy is $\Delta \mathrm{U}$ and the amount of work done by the gas is $\Delta \mathrm{W}$. The ratio $\Delta \mathrm{W}: \Delta \mathrm{U}: \Delta \mathrm{Q}$ is
$2: 3: 5$
$2: 5: 7$
$7: 5: 9$
$1: 2: 5$
Solution
Fraction of given heat energy utilised in doing external work is given by the formula,
$\begin{aligned}
& \left(\frac{\Delta \mathrm{W}}{\Delta \mathrm{Q}}\right)=\left(1-\frac{1}{\gamma}\right) \\
& \begin{aligned}
\frac{\Delta \mathrm{W}}{\Delta \mathrm{Q}} & =1-\frac{1}{\gamma} \\
& =1-\frac{1}{\left(\frac{7}{5}\right)} \quad \ldots .\left(\gamma_{\text {diatomic }}=\frac{7}{5}\right)
\end{aligned} \\
& \begin{aligned}
\frac{\Delta \mathrm{W}}{\Delta \mathrm{Q}} & =\frac{2}{7}...(i)
\end{aligned}
\end{aligned}$
The fraction of heat energy used to increase the internal energy of gas is,
$\frac{\Delta U}{\Delta Q}=\frac{1}{\gamma}=\frac{5}{7}...(ii)$ From equations (i) and (ii),
$\Delta \mathrm{W}: \Delta \mathrm{U}: \Delta \mathrm{Q}=2: 5: 7$
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