At a given temperature T , gases $\mathrm{Ne}, \mathrm{Ar}, \mathrm{Xe}$ and Kr are found to deviate from ideal gas behaviour. Their equation of state is given as $\mathrm{P}=\frac{\mathrm{RT}}{\mathrm{V}-\mathrm{b}}$ at T . Here, b is the van der Waals constant. Which gas will exhibit steeper increase in the plot of Z (compression factor) versus P ?

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · States of Matter

At a given temperature T , gases

Ne, Ar, Xe

and Kr are found to deviate from ideal gas behaviour. Their equation of state is given as

P = \frac{RT}{V - b}

at T . Here, b is the van der Waals constant. Which gas will exhibit steeper increase in the plot of Z (compression factor) versus P ?

(A)Ne
(B)Ar
(C)Xe✓ Correct
(D)Kr

Answer: (C)

Solution

Rearranging the given equation,

P (V - b) = R T

, so the compression factor is

Z = \frac{P V}{R T} = 1 + \frac{P b}{R T}

. At a fixed temperature

Z

rises linearly with

P

, with a slope set by the van der Waals constant

b

, which measures the excluded volume and grows with molecular size. Among Ne, Ar, Kr and Xe the largest atom is xenon, so it has the biggest

b

and gives the steepest rise of

Z

against

P

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option

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