A closed vessel contains a mixture of two diatomic gases A and B. Molar mass of A is 16 times that of B and mass of gas A contained in the vessel is 2 times that of B. Which of the following statements are correct?

Since both the gases are contained in the same vessel, temperature of both the gases is same.
Average KE per molecule of a diatomic gas is 512 KT. Hence, average KE per molecule of both the gases is same. Therefore, option (1) is correct.

${\mathrm{v}}_{\mathrm{rms}}=\sqrt{\frac{3\mathrm{RT}}{\mathrm{M}}}$

Hence, $\frac{{\left({\mathrm{v}}_{\mathrm{rms}}\right)}_2}{{\left({\mathrm{v}}_{\mathrm{rms}}\right)}_1}=\sqrt{\frac{{\mathrm{M}}_1}{{\mathrm{M}}_2}}=\sqrt{16}=4$

Hence, option (2) is correct.

Let molar mass of B be M, then that of A will be equal to 16M.
Let mass of gas B in the vessel be m; then that of A will be 2m. The number of moles of a gas, in the vessel will be n = m/M. Hence, number of moles of gases A and B will be

${\mathrm{n}}_1=\frac{2\mathrm{m}}{16\mathrm{M}}\text{ and }{\mathrm{n}}_2=\frac{\mathrm{m}}{\mathrm{M}}$

Hence, $\frac{{\mathrm{n}}_1}{{\mathrm{n}}_2}=\frac{1}{8}$

Hence, option (4) is correct.

Partial pressure exerted by a gas is $\mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}$

Hence, $\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}=\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1}=8$

Therefore, option (3) is also correct.