States of Matter - Result Question 98

####2. Consider the following table.

Gas a/((k Pa dmmol1) b/(dm3mol1)
A 642.32 0.05196
B 155.21 0.04136
C 431.91 0.05196
D 155.21 0.4382

a and b are van der Waals’ constants. The correct statement about the gases is

(2019 Main, 10 April I)

(a) gas C will occupy lesser volume than gas A; gas B will be lesser compressible than gas D

(b) gas C will occupy more volume than gas A; gas B will be more compressible than gas D

(c) gas C will occupy more volume than gas A; gas B will be lesser compressible than gas D

(d) gas C will occupy more volume than gas A; gas B will be lesser compressible than gas D

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Answer:

Correct Answer: 2. (c)

Solution:

  1. For 1 mole of a real gas, the van der Waals’ equation is

(p+aV2)(Vb)=RT

The constant ’ a ’ measures the intermolecular force of attraction of gas molecules and the constant ’ b ’ measures the volume correction by gas molecules after a perfectly inelastic binary collision of gas molecules.

For gas A and gas C given value of ’ b ’ is

0.05196dm3mol1. Here,

a intermolecular force of attraction

compressibility real nature

1 volume occupied 

Value of a/(kPadm6mol1) for gas A(642.32)>gasC(431.91) So, gas C will occupy more volume than gas A. Similarly, for a given value of a say 155.21kPadm6mol1 for gas B and gas D

1b intermolecular force of attraction

compressibility real nature

1 volume accupied 

b/(dm3mol1) for gas B(0.04136)<GasD(0.4382)

So, gas B will be more compressible than gas D.



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