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 $2T_0$ in a closed vessel, 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 $\frac{5}{3}{T}_0$

Let '$n$' be the total number of moles of the mixture of two gases.
Given,
Number of moles of monoatomic gas 1
$\left(n_1\right)=2\text{moles}$
Number of moles of monoatomic gas 2
$\left(n_2\right)=4\text{moles}$
Let $T$ be the temperature of the mixture. Then,
Internal energy of the mixture of two gases is given by
$U=U_1+U_2$
$\Rightarrow U=\frac{f}{2}(n_{1}R{T}_1++n_{2}R{T}_2)$
$\Rightarrow n\left(\frac{fR}{2}\right)T=\frac{fR}{2}\times n_1\times T_0+\frac{fR}{2}\times n_2\times 2T_0...\left(1\right)$
Number of moles of gas in the mixture $\left(n\right)=n_1+n_2=2+4$
Also, $f=3$ for monoatomic gas.
Using this in (1), we get
$(2+4)T=2T_0+8T_0(\because n_1=2,n_2=4)$
$\therefore T=\frac{5}{3}{T}_0$
Thus, option (a) is the correct answer.