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(B) The van der Waals' equation for $n = 1$ mole is $(p + a/V^2)(V - b) = RT$.Rearranging for the compressibility factor $Z = (pV/RT)$,we get $Z = (V/

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$b$ decreases and $a$ increases at constant temperature

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(B) The van der Waals' equation for $n = 1$ mole is $(p + a/V^2)(V - b) = RT$.Rearranging for the compressibility factor $Z = (pV/RT)$,we get $Z = (V/(V - b)) - (a/(RTV))$.To decrease $Z$ at a fixed volume $V$,the term $(V/(V - b))$ should decrease and the term $(a/(RTV))$ should increase.$1$. For $(V/(V - b))$ to decrease,the denominator $(V - b)$ must increase,which means $b$ must decrease.$2$. For $(a/(RTV))$ to increase,$a$ must increase (at constant $T$ and $V$).Therefore,$Z$ decreases if $b$ decreases and $a$ increases at constant temperature.

For one mole of a van der Waals' gas,the compressibility factor $Z = (pV/RT)$ at a fixed volume will certainly decrease,if
[Given : $a$ and $b$ are standard parameters for van der Waals' gas]

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For one mole of a van der Waals' gas,the compressibility factor $Z = (pV/RT)$ at a fixed volume will certainly decrease,if[Given : $a$ and $b$ are standard parameters for van der Waals' gas]