A closed vessel contains 10 g of an ideal gas X at 300 K, which exerts 2 atm pressure. At the same temperature, 80 g of another ideal gas Y is added to it and the pressure becomes 6 atm. The ratio of root mean square velocities of X and Y at 300 K is

For an ideal gas at fixed temperature, the pressure is proportional to the number of moles contained in a vessel of fixed volume: $$P = \dfrac{nRT}{V}$$.

Initially only gas $$X$$ is present.

$$P_1 = 2 \text{ atm}, \quad n_X = \dfrac{m_X}{M_X}, \quad m_X = 10\ \text{g}$$

After adding gas $$Y$$ (at the same temperature and in the same vessel) the total pressure becomes

$$P_2 = 6 \text{ atm}, \quad n_Y = \dfrac{m_Y}{M_Y}, \quad m_Y = 80\ \text{g}$$

Because $$T$$, $$R$$ and $$V$$ remain unchanged, the pressure ratio equals the mole ratio:

$$\dfrac{P_2}{P_1} = \dfrac{n_X + n_Y}{n_X} \Longrightarrow \dfrac{6}{2} = 3 = 1 + \dfrac{n_Y}{n_X}$$

Hence $$\dfrac{n_Y}{n_X} = 2$$ $$-(1)$$.

Substitute the expressions for the moles:

$$\dfrac{m_Y/M_Y}{m_X/M_X} = 2$$

$$\Longrightarrow \dfrac{80/M_Y}{10/M_X} = 2$$

$$\Longrightarrow \dfrac{80}{M_Y} = \dfrac{20}{M_X}$$

$$\Longrightarrow 80M_X = 20M_Y$$

$$\Longrightarrow M_Y = 4M_X$$ $$-(2)$$.

The root‐mean‐square (rms) speed of an ideal gas is given by

$$u_{\text{rms}} = \sqrt{\dfrac{3RT}{M}}$$.

For two gases at the same temperature, the ratio of their rms speeds is inversely proportional to the square root of their molar masses:

$$\dfrac{u_{\text{rms}}(X)}{u_{\text{rms}}(Y)} = \sqrt{\dfrac{M_Y}{M_X}}$$.

Using result $$(2)$$, $$M_Y = 4M_X$$, so

$$\dfrac{u_{\text{rms}}(X)}{u_{\text{rms}}(Y)} = \sqrt{4} = 2$$.

Therefore the required ratio is $$2 : 1$$.

Option D which is: $$2 : 1$$