Some Basic Concepts in Chemistry: Chemistry | JEE Main

No, because and units cancel out identically on both sides of the equation as long as both molar masses share the same unit.

600\text{K}

, the root mean square (rms) speed of gas

\text{X}

(\text{molar mass} = 40)

is equal to the most probable speed of gas

\text{Y}

90\text{K}

. The molar mass of the gas

\text{Y}

\_\_\_\_\_\_\_\text{ g mol}^{-1}

. (Nearest integer)

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Do we need to convert molar mass from

\text{g mol}^{-1}

\text{kg mol}^{-1}

No, because $R$ and units cancel out identically on both sides of the equation as long as both molar masses share the same unit.

The molar mass of gas Y is 4 g mol⁻¹.

given: For gas X: $T_X = 600\text{ K}$ , $M_X = 40\text{ g mol}^{-1}$ . For gas Y: $T_Y = 90\text{ K}$ , $M_Y = M$ . Condition: $(u_{\text{rms}})_X = (u_{\text{mp}})_Y$ .
find: Molar mass of gas Y, $M_Y$ in $\text{g mol}^{-1}$ .
strategise: Use the Maxwell-Boltzmann molecular speed relations: $u_{\text{rms}} = \sqrt{\frac{3RT}{M}}$ and $u_{\text{mp}} = \sqrt{\frac{2RT}{M}}$ . Equate the expressions for the two gases and solve for $M_Y$ .
execute: Equating the speeds: $\sqrt{\frac{3R(600)}{40}} = \sqrt{\frac{2R(90)}{M_Y}}$ . Squaring both sides and cancelling $R$ : $\frac{3 \times 600}{40} = \frac{2 \times 90}{M_Y} \implies 45 = \frac{180}{M_Y} \implies M_Y = \frac{180}{45} = 4$ .
verify: $M_Y = 4\text{ g mol}^{-1}$ is a standard integer matching helium gas, consistent with the ratio $\frac{3 \times 600}{40} = 45 = \frac{2 \times 90}{4}$ .