JEE Advanced · 2024 · 26 May 2024 · Paper 1 · States of Matter

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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

(A) $22 : 3$
(B) $22 : 1$
(C) $1 : 2$
(D) $2 : 1$ ✓ Correct

Answer: (D)

Solution

For an ideal gas,

P V = n R T

, so at constant

T

and

V

n \propto P

. Since

n = \frac{mass}{M}

, for gas X:

\frac{10}{M _{X}} \propto 2

atm. After adding Y:

\frac{10}{M _{X}} + \frac{80}{M _{Y}} \propto 6

atm. Subtracting gives

\frac{80}{M _{Y}} \propto 4

. Dividing the first by this:

\frac{M _{Y}}{8 M _{X}} = \frac{1}{2}

, so

\frac{M _{Y}}{M _{X}} = 4

. Since

v_{r m s} = \frac{3 R T}{M}

, we have

v_{r m s} \propto \frac{1}{M}

. Therefore

\frac{( v _{r m s} ) _{X}}{( v _{r m s} ) _{Y}} = \frac{M _{Y}}{M _{X}} = 4 = 2

, giving the ratio

2 : 1

JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Single correct option

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JEE Advanced · 2024 · 26 May 2024 · Paper 1 · States of Matter

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