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Question

(B) For an ideal gas,$PV = nRT$. Given $n = 1 \, mol$,$V = 22.4 \, L$,and $T = 273 \, K$,the ideal pressure $P_{ideal} = \frac{nRT}{V} = \frac{1 	ime

Vedclass · Accepted Answer

If compressibility factor $(Z) > 1$,then $P$ will be greater than $1 \, atm$.

Vedclass · Answer

(B) For an ideal gas,$PV = nRT$. Given $n = 1 \, mol$,$V = 22.4 \, L$,and $T = 273 \, K$,the ideal pressure $P_{ideal} = \frac{nRT}{V} = \frac{1 	imes 0.0821 	imes 273}{22.4} \approx 1 \, atm$.By definition,the compressibility factor is $Z = \frac{PV}{nRT}$.Therefore,$P = \frac{Z 	imes nRT}{V} = Z 	imes P_{ideal}$.If $Z > 1$,then $P > P_{ideal}$,which means $P > 1 \, atm$.

The correct option regarding a container containing $1 \, mol$ of a gas in a $22.4 \, L$ container at $273 \, K$ is:

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