At low pressure,the Van der Waals equation is | vedclass

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(A) The Van der Waals equation for $1 \ mol$ of gas is: $(P + \frac{a}{V_m^2})(V_m - b) = RT$. At low pressure,$V_m$ is very large,so $b$ is negligibl

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$Z = \frac{PV_m}{RT} = 1 - \frac{a}{V_m RT}$

Vedclass · Answer

(A) The Van der Waals equation for $1 \ mol$ of gas is: $(P + \frac{a}{V_m^2})(V_m - b) = RT$. At low pressure,$V_m$ is very large,so $b$ is negligible compared to $V_m$ $(V_m - b \approx V_m)$. The equation becomes: $(P + \frac{a}{V_m^2})V_m = RT$. Expanding this: $PV_m + \frac{a}{V_m} = RT$. Dividing by $RT$: $\frac{PV_m}{RT} + \frac{a}{V_m RT} = 1$. Therefore,the compressibility factor $Z = \frac{PV_m}{RT} = 1 - \frac{a}{V_m RT}$.

$P \text{ (bar)}$	$\frac{PV}{RT}$
$1$	$1$
$2$	$1$
$3$	$1$
$3.1$	$1.2$
$3.5$	$1.4$
$4.0$	$1.5$

At low pressure,the Van der Waals equation is reduced to:

Similar Questions

Under which of the following conditions does a real gas behave most like an ideal gas?

For a gas,the deviation from ideal behavior is maximum at:

The following data is obtained for one mole of a gas. The gas deviates from ideal behavior in the pressure (in bar) range:
$P \text{ (bar)}$ $\frac{PV}{RT}$
$1$ $1$
$2$ $1$
$3$ $1$
$3.1$ $1.2$
$3.5$ $1.4$
$4.0$ $1.5$

The graph between $P$ and $V$ below the critical temperature is:

$A$ gas deviates most from ideal behavior when it is subjected to

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