The pressure exerted by 6.0 ~g of methane gas in a 0.03 ~m^3 vessel at 129^ C is (Atomic masses : C=12.01, H=1.01 and R=8.314 JK^ -1 ~mol^ -1 )

Correct answer: C. 41648 ~Pa Given, volume, V=0.03 ~m^3 temperature, T=129+273=402 ~K mass of methane, w=6.0 ~g mol. mass of methane, M=12.01+4 × 1.01=16.05 From, ideal gas equation, aligned & pV=nRT & p=(6)/(16.05) × (8.314 × 402)/(0.03)=41648 ~Pa aligned

NEET 2010

Exam: NEET
Year: 2010
Subject: Chemistry
Chapter: States of Matter

NEET 2010 · Chemistry · States of Matter

The pressure exerted by 6.0 ~g of methane gas in a 0.03 ~m^3 vessel...

The pressure exerted by

6.0 \mathrm{~g}

of methane gas in a

0.03 \mathrm{~m}^3

vessel at

129^{\circ} \mathrm{C}

is (Atomic masses :

\mathrm{C}=12.01, \mathrm{H}=1.01

and

\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}

)

Options

A $215216 \mathrm{~Pa}$
B $13409 \mathrm{~Pa}$
C $41648 \mathrm{~Pa}$
D $31684 \mathrm{~Pa}$

Correct answer

C. $41648 \mathrm{~Pa}$

Step-by-step solution

Given, volume,

\mathrm{V}=0.03 \mathrm{~m}^3

temperature,

\mathrm{T}=129+273=402 \mathrm{~K}

mass of methane,

w=6.0 \mathrm{~g}

mol. mass of methane,

M=12.01+4 \times 1.01=16.05

From, ideal gas equation,

\begin{aligned}& \mathrm{pV}=\mathrm{nRT}\\ & \mathrm{p}=\frac{6}{16.05} \times \frac{8.314 \times 402}{0.03}=41648 \mathrm{~Pa} \end{aligned}

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