Two vessels of volumes 16.4 L and 5 L contain two ideal gases of molecular existence at the respective temperature of 27^ ^ C and 227^ ^ C and exert 1.5 and 4.1 atmospheres respectively. The ratio of the number of

Correct answer: A. 2 Given conditions V_ 1 =16.4 L, V_ 2 =5 LP_ 1 =1.5 atm, P_ 2 =4.1 atmT_ 1 =273+27=300 KT_ 2 =273+227=500 K Applying gas equation, P_ 1 V_ 1 P_ 2 V_ 2 = n_ 1 T_ 1 n_ 2 T_ 2 n_ 1 n_ 2 = P_ 1 V_ 1 T_ 1 P_ 2 V_ 2 T_ 2 (1.5 × 16.4 × 500)/(4.1 × 5 × 300)=(2)/(1)

BITSAT 2020

Exam: BITSAT
Year: 2020
Subject: Chemistry
Chapter: States of Matter

BITSAT 2020 · Chemistry · States of Matter

Two vessels of volumes 16.4 L and 5 L contain two ideal gases of...

Two vessels of volumes

16.4 \mathrm{ L}

and

5 \mathrm{ L}

contain two ideal gases of molecular existence at the respective temperature of

27^{{}^{\circ}} \mathrm{C}

and

227^{{}^{\circ}} \mathrm{C}

and exert 1.5 and 4.1 atmospheres respectively. The ratio of the number of molecules of the former to that of the later is

Options

A2
B1
C $\frac{1}{2}$
D $\frac{1}{3}$

Correct answer

A. 2

Step-by-step solution

Given conditions

\mathrm{V}_{1}=16.4 \mathrm{ L}, \mathrm{ V}_{2}=5 \mathrm{ L}\mathrm{P}_{1}=1.5 \mathrm{ atm}, \mathrm{P}_{2}=4.1 \mathrm{ atm}\mathrm{T}_{1}=273+27=300 \mathrm{ K}\mathrm{T}_{2}=273+227=500 \mathrm{ K}

Applying gas equation,

\frac{\mathrm{P}_{1} \mathrm{ V}_{1}}{\mathrm{P}_{2} \mathrm{ V}_{2}}=\frac{\mathrm{n}_{1} \mathrm{ T}_{1}}{\mathrm{n}_{2} \mathrm{ T}_{2}}\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\frac{\mathrm{P}_{1} \mathrm{ V}_{1} \mathrm{ T}_{1}}{\mathrm{P}_{2} \mathrm{ V}_{2} \mathrm{ T}_{2}}\therefore \frac{1.5 \times 16.4 \times 500}{4.1 \times 5 \times 300}=\frac{2}{1}

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