Question

Gases $X,Y,Z,P$ and $Q$ have their van der Waals constants ${}^{\prime }{a}^{\prime }$ and ${}^{\prime }{b}^{\prime }$ (in CGS units) as shown below:
$\begin{array}{cccccc}& X& Y& Z& P& Q\\ a& 6& 6& 20& 0.05& 30\\ b& 0.025& 0.15& 0.1& 0.02& 0.2\end{array}$
The gas with the highest critical temperature is:

• A. $P$
• B. $Q$
• C. $Y$
• D. $X$

Answer 1

The correct option is D $X$

For any gas, critical temperature $\left(T_c\right)$ is the temperature at or above which the gas cannot be liquefied, $T_c=\frac{8a}{27Rb}$, where a ,b are van der Waals' constant and R is the gas constant.
$\therefore$ Greater the value of $\left(\frac{a}{b}\right)$, more will be the critical temperature of a gas.
a) For gas $P$, value of $\frac{a}{b}$ = $\frac{0.05}{0.02}=2.5$
b) For gas $Q$, value of $\frac{a}{b}$ = $\frac{30}{0.2}=150$
c) For gas $Y$, value of $\frac{a}{b}$ = $\frac{6}{0.15}=40$
d) For gas $Z$, value of $\frac{a}{b}$ = $\frac{20}{0.1}=200$
e) For gas $X$, value of $\frac{a}{b}$ = $\frac{6}{0.025}=240$
For gas $X$, $\left(\frac{a}{b}\right)$ has the maximum value. Hence it has the highest critical temperature $T_c$ .

Theory:

Critical Point (Point of inflection):

At this point on critical isotherm, gas and liquid are at equilibrium.

At point J (the tip of the dome) :

$\left(\frac{dP}{dV}\right)=\left(\frac{d^{2}P}{d{V}^2}\right)=0$

At Higher Temperature

$T>T_c$

Gas can not be liquefied for any $T>T_c$

Critical Temperature $\left(T_c\right)$ : The maximum temperature at which a gas can be liquefied by the application of pressure alone.

Critical Pressure $\left(P_c\right)$ :The minimum pressure required to liquefy a gas at critical temperature $\left(T_c\right)$

Critical Volume $\left(V_c\right)$ : The volume occupied by one mole of the gas at critical temperature $\left(T_c\right)$, and critical pressure $\left(P_c\right)$

Critical constants

Critical Temperature $\left(T_c\right)$, Critical Pressure $\left(P_c\right)$ and Critical Volume $\left(V_c\right)$ are called critical constants.

Significance of Critical Temperature $\left(T_c\right)$

$T_c\propto \text{Intermolecular attractions}\propto \text{Ease of liquefaction}$

Critical Constants in Terms of van der Waals Constants

$V_c=3b$

$P_c=\frac{a}{27\times b^2}$

$T_c=\frac{8a}{27Rb}$

Compressibility Factor and Critical Constants for every gas

$Z_c=\frac{P_c{V}_c}{R{T}_c}=\frac{27ab}{\left(9b\right)\left(8a\right)}=\frac{3}{8}=0.375$