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Question

Question: The perfect gas equation for \(4\)grams of hydrogen gas is: A. \(PV = RT\) B. \(PV = 2RT\) C. ...

The perfect gas equation for 44grams of hydrogen gas is:
A. PV=RTPV = RT
B. PV=2RTPV = 2RT
C. PV=12RTPV = \dfrac{1}{2}RT
D. PV=4RTPV = 4RT

Explanation

Solution

The general equation of perfect gas is to be used. From the given values in question, after calculating the number of moles of the equation, the answer can be obtained.
By using formula: II - ideal perfect gas equation
PV=nRTPV = nRT
Where PPindicates volume ,nn represents no. of moles,RR is a gas constant and TT is temperature.

Complete step by step answer:
We know that number of moles of a gas is given by
n=givenmassmolarmass=mMn = \dfrac{{given\,mass}}{{molar\,\,mass}} = \dfrac{m}{M}
Where mmsignifies available mass of gas and MMsignifies the molar mass of the gas. For hydrogen gas,
Molar mass, M=2gmolM = 2g\,mol
And given mass ,m=4gm = 4g
So, no of moles 42=2moles\dfrac{4}{2} = 2\,moles
Now, the ideal gas equation is given by PV=nRTPV = nRT
Where PPdenotes pressure, VV is volume, nn is no. of moles, RR is gas constant, R=3.314R = 3.314 and TT is temperature.
For 4g4g of hydrogen gas,
No. of moles, n=2n = 2
So, the ideal gas equation becomes
PV=(2)RTPV = (2)RT
PV=2RTPV = 2RT
Hence, the correct option is B.

Note: This can also be solved by unitary method to find numbers of moles of hydrogen gas. Molar mass of hydrogen gas =2g/mol = 2g/mol. This means,
Corresponding 2g2gof hydrogen gas, no. of moles =1 = 1mole
Corresponding to 1g1gof hydrogen gas, no. of moles =12 = \dfrac{1}{2}mole
Corresponding to 4g4gof hydrogen gas, no. of moles =42=2moles = \dfrac{4}{2} = 2moles
So, n=2n = 2moles,
Also, no gas is perfectly ideal as ideal gas has no interactions within its particles and thus no kinetic energy of particles which is not possible.