Question

Two vessels connected by a valve of negligible volume. One container (I) has 2.8 g of $N_2$ at temperature $T_1(K)$. The other container (II) is completely evacuated. The container (I) is heated to $T_2(K)$ while container (II) is maintained at $T_2/3(K)$. volume of vessel (I) is half that of vessel (II). If the valve is opened then what is the mass ratio of $N_2$ in both vessel $(W_I/W_{II})$?

• A. 1 : 2
• B. 1 : 3
• C. 1 : 6
• D. 3 : 1

Answer 1

Answer: (C)

1 : 6

Answer 2

Equalizing pressure in both vessels gives:

$$P = \frac{n_I R T_2}{V} = \frac{n_{II} R \,(T_2/3)}{2V}$$

Canceling $R$ and simplifying:

$$\frac{n_I T_2}{V} = \frac{n_{II} T_2}{6V} \quad\Longrightarrow\quad n_I = \frac{n_{II}}{6}$$

Thus the mass ratio is

$$\frac{W_I}{W_{II}} = \frac{n_I}{n_{II}} = \frac{1}{6}.$$