What is the compressibility factor $(Z)$ for $0.02$ mole of a van der waal's gas at pressure of $0.1 a t m$ . Assume the size of gas molecule is negligible.
(Given: $R T = 20 L a t m m o l^{- 1}$ and $a = 1000 a t m L^{- 2} m o l^{- 1}$ )

2

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Solution

The correct option is D $0.5$
As the size of the gas molecule is negligble, then the volume correction $b$ is also neglible.
$(0.1 + \frac{1000 \times (0.02)^{2}}{V^{2}}) V = 20 \times 0.02 = 0.1 V^{2} - 0.4 V + 0.4 = 0 \Rightarrow V^{2} - 4 V + 4 = 0 \Rightarrow V = 2 L Z = \frac{P V}{n R T} = \frac{0.1 \times 2}{0.02 \times 20} = 0.5$

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What is the compressibility factor (Z) for 0.02 mole of a van der waal's gas at pressure of 0.1 atm. Assume the size of gas molecule is negligible. (Given: RT=20 L atm mol−1 and a=1000 atm L−2 mol−1)

The correct option is D 0.5As the size of the gas molecule is negligble, then the volume correction b is also neglible. (0.1+1000×(0.02)2V2)V=20×0.02=0.1V2−0.4V+0.4=0⇒V2−4V+4=0⇒V=2 LZ=PVnRT=0.1×20.02×20=0.5

What is the compressibility factor $(Z)$ for $0.02$ mole of a van der waal's gas at pressure of $0.1 a t m$ . Assume the size of gas molecule is negligible.
(Given: $R T = 20 L a t m m o l^{- 1}$ and $a = 1000 a t m L^{- 2} m o l^{- 1}$ )