JEE Advanced 2023 Paper 1, Chemistry Section 3 Q2: Deviation From Ideal Gas Behaviour

JEE Advanced2023Paper 1Chemistry Section 3

A gas has a compressibility factor of $0.5$ and a molar volume of $0.4 dm^{3} mol^{- 1}$ at a temperature of $800$ K and pressure $x$ atm. If it shows ideal gas behaviour at the same temperature and pressure, the molar volume will be $y$ dm $^{3}$ mol $^{- 1}$ . The value of $x / y$ is ___.

[Use: Gas constant, $R = 8 \times 1 0^{- 2} L atm K^{- 1} mol^{- 1}$ ]

Your answer

Solution

Find the pressure x. Using $Z = \frac{P V _{m}}{R T}$ : $0.5 = \frac{x \cdot 0.4}{( 8 \times 1 0 ^{- 2} ) ( 800 )} = \frac{0.4 x}{64} \Rightarrow x = 80 atm .$

Ideal molar volume at the same T and P. $y = \frac{R T}{P} = \frac{( 8 \times 1 0 ^{- 2} ) ( 800 )}{80} = 0.8 dm^{3} mol^{- 1} .$

Ratio. $\frac{x}{y} = \frac{80}{0.8} = 100$ .

Answer: 100.

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JEE Advanced 2023 Paper 1, Chemistry Section 3 Q2: Deviation From Ideal Gas Behaviour

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