Question

If $2$ moles of an ideal monoatomic gas at temperature $T_0$ is mixed with $4$ moles of another ideal monoatomic gas at temperature $2T_0$, then the temperature of the mixture is

• A. $\frac{5}{3}{T}_0$
• B. $\frac{3}{2}{T}_0$
• C. $\frac{4}{3}{T}_0$
• D. $\frac{5}{4}{T}_0$

Answer 1

The correct option is A $\frac{5}{3}{T}_0$

For the internal energy of mixture,
$U_{mix}=\frac{(n_1+n_2)fRT}{2}$
Degree of freedom $f$ will be same for both gases and the mixture as gases mixed are monoatomic.
For the mixture we can apply,
$U_{mix}=U_1+U_2$
$\Rightarrow \frac{(n_1+n_2)fRT}{2}=\frac{n_{1}fR{T}_0}{2}+\frac{n_{2}fR\times 2T_0}{2}$
$\Rightarrow (n_1+n_2)T=n_1{T}_0+2n_2{T}_0$
$\Rightarrow (2+4)T=2T_0+8T_0$
$\therefore T=\frac{10T_0}{6}=\frac{5}{3}{T}_0$