A container has two chambers of volumes V_1=2 litres and V_2=3...

A container has two chambers of volumes

V_1=2

litres and

V_2=3

litres separated by a partition made of a thermal insulator. The chambers contains

n_1=5

and

n_2=4

moles of ideal gas at pressures

p_1=1 \mathrm{~atm}

and

p_2=2 \mathrm{~atm}

, respectively. When the partition is removed, the mixture attains an equilibrium pressure of :

Options

B. 1.6 atm

By energy conservation

\begin{aligned} & \mathrm{E}_1+\mathrm{E}_2=\mathrm{E}_{\text{mix }} \\ & \frac{\mathrm{n}_1 \mathrm{f}_1 \mathrm{RT}_1}{2}+\frac{\mathrm{n}_2 \mathrm{f}_2 \mathrm{RT}_2}{2}=\frac{\left(\mathrm{n}_1 \times \mathrm{n}_2\right) \mathrm{fRT}_{\text{mix }}}{2} \\ & \Rightarrow \frac{3}{2} \mathrm{P}_1 \mathrm{ V}_1+\frac{3}{2} \mathrm{P}_2 \mathrm{ V}_2=\frac{3}{2} \mathrm{P}_{\text{mix }}\left(\mathrm{V}_1+\mathrm{V}_2\right) \\ & \frac{3}{2} \times 1 \times 2+\frac{3}{2} \times 2 \times 3=\frac{3}{2} \times \mathrm{P}_{\text{mix }}(5) \\ & \therefore \mathrm{P}_{\text{mix }}=\frac{8}{5} \\ & \mathrm{P}_{\text{mix }}=1.6 \mathrm{ atm} \end{aligned}