At low pressure,Vander Waal's equation is reduced to $\left[ P + \frac{a}{V^2} \right]V = RT$. The compressibility factor $(Z)$ can be given as
- A$1 - \frac{a}{RTV}$
- B$1 - \frac{RTV}{a}$
- C$1 + \frac{a}{RTV}$
- D$1 + \frac{RTV}{a}$
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$A$ real gas most closely approaches the behaviour of an ideal gas at
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View SolutionUnder which of the following conditions does a real gas behave most like an ideal gas?
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View SolutionConsider the following table:
Gas $a / (kPa \cdot dm^6 \cdot mol^{-2})$ $b / (dm^3 \cdot mol^{-1})$ $A$ $642.32$ $0.05196$ $B$ $155.21$ $0.04136$ $C$ $431.91$ $0.05196$ $D$ $155.21$ $0.4382$
$a$ and $b$ are van der Waals constants. The correct statement about the gases is:
| Gas | $a / (kPa \cdot dm^6 \cdot mol^{-2})$ | $b / (dm^3 \cdot mol^{-1})$ |
|---|---|---|
| $A$ | $642.32$ | $0.05196$ |
| $B$ | $155.21$ | $0.04136$ |
| $C$ | $431.91$ | $0.05196$ |
| $D$ | $155.21$ | $0.4382$ |
$a$ and $b$ are van der Waals constants. The correct statement about the gases is:
DifficultJEE MAIN 2019
View Solution$A$ gas at $350 \; K$ and $15 \; bar$ has a molar volume $20$ percent smaller than that for an ideal gas under the same conditions. The correct option about the gas and its compressibility factor $(Z)$ is
MediumNEET 2019
View SolutionAssertion : Compressibility factor $(Z)$ for non-ideal gases can be greater than $1$.
Reason : Non-ideal gases always exert higher pressure than expected.
Reason : Non-ideal gases always exert higher pressure than expected.
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