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

Correct answer: D. 2: 1 For Ideal Gas PV=nRT n P at constant T \& ~V . mole = Mass Molar mass For gas X: 10 M_ X 2 ~atm (1) For gas X \& Y: 10 M_ X + 80 M_ Y 6 ~atm (2) From (2)-(1) 80 M_ y 4 (3) On dividing (1) by (3) M_ Y 8 M_ X =(1)/(2) M_ Y M_ X =4 (4) V_ rms = 3 RT M v_ rms 1 M (V_ ms )_ X (V_ ms )_ Y = M_ Y M_ x = (4)/(1) =(2)/(1)

JEE Advanced 2024Paper 1

Exam: JEE Advanced
Year: 2024
Paper: Paper 1
Subject: Chemistry
Chapter: States of Matter

JEE Advanced 2024 · Paper 1 · Chemistry · States of Matter

A closed vessel contains 10 ~g of an ideal gas X at 300 ~K , which...

A closed vessel contains

10 \mathrm{~g}

of an ideal gas

\mathbf{X}

300 \mathrm{~K}

, which exerts 2 atm pressure. At the same temperature,

80 \mathrm{~g}

of another ideal gas

\mathbf{Y}

is added to it and the pressure becomes

6 \mathrm{~atm}

. The ratio of root mean square velocities of

\mathbf{X}

and

\mathbf{Y}

300 \mathrm{~K}

Options

A $2 \sqrt{2}: \sqrt{3}$
B $2 \sqrt{2}: 1$
C $1: 2$
D $2: 1$

Correct answer

D. $2: 1$

Step-by-step solution

For Ideal Gas

\mathrm{PV}=\mathrm{nRT}\therefore \mathrm{n} \propto \mathrm{P}

at constant

\mathrm{T} \& \mathrm{~V}

\because \quad

mole

=\frac{\text{ Mass }}{\text{ Molar mass }}

For gas

\mathrm{X}: \frac{10}{\mathrm{M}_{\mathrm{X}}} \propto 2 \mathrm{~atm}\quad\ldots

(1) For gas

\mathrm{X} \& \mathrm{Y}: \frac{10}{\mathrm{M}_{\mathrm{X}}}+\frac{80}{\mathrm{M}_{\mathrm{Y}}} \propto 6 \mathrm{~atm} \ldots

(2) From

(2)-(1) \frac{80}{\mathrm{M}_{\mathrm{y}}} \propto 4 \ldots

(3) On dividing (1) by (3)

\frac{\mathrm{M}_{\mathrm{Y}}}{8 \mathrm{M}_{\mathrm{X}}}=\frac{1}{2}\therefore \quad \frac{\mathrm{M}_{\mathrm{Y}}}{\mathrm{M}_{\mathrm{X}}}=4 \ldots

(4)

\because \mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \Rightarrow \mathrm{v}_{\mathrm{rms}} \propto \frac{1}{\sqrt{\mathrm{M}}} \therefore \quad \frac{\left(\mathrm{V}_{\mathrm{ms}}\right)_{\mathrm{X}}}{\left(\mathrm{V}_{\mathrm{ms}}\right)_{\mathrm{Y}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{Y}}}{\mathrm{M}_{\mathrm{x}}}}=\sqrt{\frac{4}{1}}=\frac{2}{1}

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