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A gas mixture contains monoatomic and diatomic molecules of 2 moles each. The mixture has a total internal energy of (symbols have usual meanings)
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Verified Answer
The correct answer is:
$8 R T$
Total internal energy of a gas is given as
$U=\frac{n}{2} f R T$
where, $n=$ number of moles
and $f=$ degree of freedom.
Given, $n_{\text {diatomic }}=n_{\text {monoatomic }}=2$
As, $f_{\text {monoatomic }}=3$
$\begin{aligned}
f_{\text {diatomic }} &=5 \\
\Rightarrow U_{\text {monoatomic }} &=\frac{2}{2} \times 3 R T=3 R T \\
U_{\text {diatomic }} &=\frac{2}{2} \times 5 R T=5 R T
\end{aligned}$
$\therefore$ Total internal energy of the mixture of gases,
$\begin{aligned}
U_{\text {total }} &=U_{\text {monoatomic }}+U_{\text {diatomic }} \\
&=3 R T+5 R T=8 R T
\end{aligned}$
$U=\frac{n}{2} f R T$
where, $n=$ number of moles
and $f=$ degree of freedom.
Given, $n_{\text {diatomic }}=n_{\text {monoatomic }}=2$
As, $f_{\text {monoatomic }}=3$
$\begin{aligned}
f_{\text {diatomic }} &=5 \\
\Rightarrow U_{\text {monoatomic }} &=\frac{2}{2} \times 3 R T=3 R T \\
U_{\text {diatomic }} &=\frac{2}{2} \times 5 R T=5 R T
\end{aligned}$
$\therefore$ Total internal energy of the mixture of gases,
$\begin{aligned}
U_{\text {total }} &=U_{\text {monoatomic }}+U_{\text {diatomic }} \\
&=3 R T+5 R T=8 R T
\end{aligned}$
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