The pressure exerted by $6.0 \,g$ of methane gas in a $0.03\, m ^{3}$ vessel at $129^{\circ} \,C$ is (Atomic masses: $C =12.01, \,H =1.01$ and $\left.R=8.314\, JK ^{-1} \,mol ^{-1}\right)$
- 215216 Pa
- 13409 Pa
- 41648 Pa
- 31684 Pa
The Correct Option is C
Solution and Explanation
temperature, $T=129+273=402\, K$
mass of methane, $W=6.0\, g$
mol mass of methane, $M=12.01+4 \times 1.01$
$=16.05$
From, ideal gas equation,
$p V=n R T$
$p=\frac{6}{16.05} \times \frac{8.314 \times 402}{0.03} $
$=41648\, Pa$
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Concepts Used:
Ideal Gas Equation
An ideal gas is a theoretical gas composed of a set of randomly-moving point particles that interact only through elastic collisions.
What is Ideal Gas Law?
The ideal gas law states that the product of the pressure and the volume of one gram molecule of an ideal gas is equal to the product of the absolute temperature of the gas and the universal gas constant.
PV=nRT
where,
P is the pressure
V is the volume
n is the amount of substance
R is the ideal gas constant
Ideal Gas Law Units
When we use the gas constant R = 8.31 J/K.mol, then we have to plug in the pressure P in the units of pascals Pa, volume in the units of m3 and the temperature T in the units of kelvin K.