JEE Advance - Chemistry (2023 - Paper 1 Online - No. 9)
A gas has a compressibility factor of 0.5 and a molar volume of $0.4 ~\mathrm{dm}^3 \mathrm{~mol}^{-1}$ at a temperature of $800 \mathrm{~K}$ and pressure $\mathbf{x}$ atm. If it shows ideal gas behaviour at the same temperature and pressure, the molar volume will be $\mathbf{y} ~\mathrm{dm}^3 \mathrm{~mol}^{-1}$. The value of $\mathbf{x} / \mathbf{y}$ is __________.
[Use: Gas constant, $\mathrm{R}=8 \times 10^{-2} \mathrm{~L}$ atm $\mathrm{K}^{-1} \mathrm{~mol}^{-1}$ ]
[Use: Gas constant, $\mathrm{R}=8 \times 10^{-2} \mathrm{~L}$ atm $\mathrm{K}^{-1} \mathrm{~mol}^{-1}$ ]
Answer
100
Explanation
For gas : $\mathrm{Z}=0.5, \mathrm{~V}_{\mathrm{m}}=0.4 \mathrm{~L} / \mathrm{mol}$
$$ \begin{aligned} & \mathrm{T}=800 \mathrm{~K}, \mathrm{P}=\mathrm{X} \text { atm. } \\\\ \Rightarrow & \mathrm{Z}=\frac{\mathrm{PV}_{\mathrm{m}}}{\mathrm{RT}} \\\\ \Rightarrow & \frac{\mathrm{X}(0.4)}{0.08 \times 800}=0.5 \\\\ \Rightarrow & \mathrm{X}=80 \end{aligned} $$
For ideal gas, $\mathrm{PV}_{\mathrm{m}}=\mathrm{RT}$
$$ \Rightarrow \mathrm{V}_{\mathrm{m}}=\frac{\mathrm{RT}}{\mathrm{P}}=\frac{0.08 \times 800}{80}=0.8 \mathrm{~L} \mathrm{~mol}^{-1}=\mathrm{y} $$
Then, $\frac{x}{y}=\frac{80}{0.8}=100$.
$$ \begin{aligned} & \mathrm{T}=800 \mathrm{~K}, \mathrm{P}=\mathrm{X} \text { atm. } \\\\ \Rightarrow & \mathrm{Z}=\frac{\mathrm{PV}_{\mathrm{m}}}{\mathrm{RT}} \\\\ \Rightarrow & \frac{\mathrm{X}(0.4)}{0.08 \times 800}=0.5 \\\\ \Rightarrow & \mathrm{X}=80 \end{aligned} $$
For ideal gas, $\mathrm{PV}_{\mathrm{m}}=\mathrm{RT}$
$$ \Rightarrow \mathrm{V}_{\mathrm{m}}=\frac{\mathrm{RT}}{\mathrm{P}}=\frac{0.08 \times 800}{80}=0.8 \mathrm{~L} \mathrm{~mol}^{-1}=\mathrm{y} $$
Then, $\frac{x}{y}=\frac{80}{0.8}=100$.
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