Question

van der Waals' constants of two gases $X$ and $Y$ are as given:
$\begin{array}{ccc}& a\left(L^{2}atm{\text{mol}}^{-2}\right)& b\left({\text{L mol}}^{-1}\right)\\ \text{Gas}X& 5.6& 0.065\\ \text{Gas}Y& 5.1& 0.012\end{array}$
Which option is true?

• A. $T_c\left(X\right)>T_c\left(Y\right)$
• B. $T_c\left(X\right)=T_c\left(Y\right)$
• C. $V_c\left(X\right)>V_c\left(Y\right)$
• D. $V_c\left(Y\right)>V_c\left(X\right)$

Answer 1

The correct option is C $V_c\left(X\right)>V_c\left(Y\right)$

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 Waal's constants and R is the gas constant.
$\therefore$ Greater the value of $\left(\frac{a}{b}\right)$, more will be the critical temperature of the gas.
For gas $X$, value of $\frac{a}{b}=\frac{5.6}{0.065}=86.15$
For gas $Y$, value of $\frac{a}{b}=\frac{5.1}{0.012}=425$
So value of $\frac{a}{b}$ for $Y$ is greater than $X$. Hence the critical temperature for $Y$ will be greater than $X$.
The volume occupied by a mole of a gas in its critical state is known as critical volume $V_c$.
$V_c=3b$, where b is van der waals' constant.
$\begin{array}{ccc}& \text{Gas}X& \text{Gas}Y\\ V_c=3b& 3\times 0.065=0.195& 3\times 0.012=0.036\end{array}$
So the critical volume of $X$ is greater than that of $Y$.