For a gas at STP,the compressibility factor i | vedclass

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(C) The compressibility factor $Z$ is defined as $Z = \frac{PV_m}{RT}$.For an ideal gas,$Z = 1$ and the molar volume $V_m$ at $STP$ is $22.4 \ dm^3$.G

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Greater than $22.4 \ dm^3$

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(C) The compressibility factor $Z$ is defined as $Z = \frac{PV_m}{RT}$.For an ideal gas,$Z = 1$ and the molar volume $V_m$ at $STP$ is $22.4 \ dm^3$.Given that $Z > 1$,the gas shows positive deviation from ideal behavior,meaning the repulsive forces are dominant.Since $Z = \frac{V_{m, real}}{V_{m, ideal}}$,if $Z > 1$,then $V_{m, real} > V_{m, ideal}$.Therefore,the molar volume $V_m$ will be greater than $22.4 \ dm^3$.

For a gas at $STP$,the compressibility factor is greater than $1$. What will be the value of its molar volume $(V_m)$?

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Assertion : Compressibility factor $(Z)$ for non-ideal gases can be greater than $1$.
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