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1 mole of a real gas is kept at high pressure of 100 bar at $300 \mathrm{~K}$. If van der Waals constant $b$ is $0.005 \mathrm{~L} / \mathrm{mol}$, what are the values of compressibility factor $\mathrm{Z}$ of the gas and $\%$ deviation of volume from ideality?
$\begin{array}{llc} & \mathbf{Z} & \text { \% Deviation } \end{array}$
Options:
$\begin{array}{llc} & \mathbf{Z} & \text { \% Deviation } \end{array}$
Solution:
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Verified Answer
The correct answer is:
$1.02 \quad 2$
$\mathrm{P}(\mathrm{V}-\mathrm{nb})=\mathrm{nRT} ; \mathrm{n}=1$ mole, $\mathrm{T}=300 \mathrm{~K}, \mathrm{P}=100 \mathrm{bar}$
$\therefore \quad 100(\mathrm{~V}-0.005)=1 \times 0.082 \times 1.013 \times 300$
or, $\mathrm{V}=(0.249+0.005) \mathrm{L}$
or, $\mathrm{V}=0.254 \mathrm{~L}$
$\mathrm{V}_{\text {ideal }}=0.254 \mathrm{~L}, \mathrm{~V}_{\text {real }}=0.249 \mathrm{~L}$
$\therefore \quad \mathrm{Z}=\frac{\mathrm{V}_{\text {ideal }}}{\mathrm{V}_{\text {real }}}=\frac{0.254}{0.249}=1.02$
$\%$ Deviation of volume $=\frac{0.005}{0.254} \times 100 \simeq 2$
$\therefore \quad 100(\mathrm{~V}-0.005)=1 \times 0.082 \times 1.013 \times 300$
or, $\mathrm{V}=(0.249+0.005) \mathrm{L}$
or, $\mathrm{V}=0.254 \mathrm{~L}$
$\mathrm{V}_{\text {ideal }}=0.254 \mathrm{~L}, \mathrm{~V}_{\text {real }}=0.249 \mathrm{~L}$
$\therefore \quad \mathrm{Z}=\frac{\mathrm{V}_{\text {ideal }}}{\mathrm{V}_{\text {real }}}=\frac{0.254}{0.249}=1.02$
$\%$ Deviation of volume $=\frac{0.005}{0.254} \times 100 \simeq 2$
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