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If $1$ mole of an ideal monoatomic gas is mixed with 2 moles of an ideal diatomic gas, what is the temperature of the given mixture, given that the monoatomic gas is at temperature $T_0$ and diatomic gas is at temperature $2T_0$?

$\begin{array}{1 1} 1.33\;T_0 \\ 1.77 \;T_0 \\ 1.5\;T_0 \\ 1.25\;T_0 \end{array}$

1 Answer

Since the final internal energy will be the sum of the internal energies of individual gases, $u = u_1 + u_2$,
$\Rightarrow \large \frac{(n_1 + n_2) f R T}{2} $$ = \large \frac{n_1f_1 R T_0}{2} $$+ \large\frac{n_2f_2R 2T_0}{2}$, where $T$ is the temperature of the mixture and $f$ is the degrees of freedom of the mixture.
$\Rightarrow 3 f T = 3 T_0 + 20T_0$
Now, $f = \large\frac{n_1f_1+n_2f_2}{n_1+n_2} = \frac{13}{3}$
Substituting, we get $13T = 23T_0 \rightarrow T = \large\frac{23}{13}$$T_0 = 1.77 T_0$
answered Jan 22, 2014 by balaji.thirumalai

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