Question

Match the Van der Waal's for one $\text{mole}$ of gas given in column $2$ to the conditions in column $1$.
$$\begin{array}{cccc}& \text{Column 1}& & \text{Column 2}\\ A& \text{At low pressure}& i& P(V-b)=RT\\ B& \text{At high pressure}& ii& (P+\frac{a}{V^2})V=RT\\ C& \text{At low pressure and high temperature}& iii& PV=RT\\ D& \text{At room temperature and pressure}& iv& (P+\frac{a}{V^2})(V-b)=RT\end{array}$$

• A. $A-\left(i\right);B-\left(ii\right);C-\left(iii\right);D-\left(iv\right)$
• B. $A-\left(ii\right);B-\left(i\right);C-\left(iii\right);D-\left(iv\right)$
• C. $A-\left(iv\right);B-\left(iii\right);C-\left(ii\right);D-\left(i\right)$
• D. $A-\left(iii\right);B-\left(ii\right);C-\left(i\right);D-\left(iv\right)$

Answer 1

The correct option is B $A-\left(ii\right);B-\left(i\right);C-\left(iii\right);D-\left(iv\right)$

We know that,
At room temperature and pressure, generally Van der Waal's equation for one $\text{mole}$ of gas is
$(P+\frac{a}{V^2})(V-b)=RT$
At low pressure,
$P\downarrow \Rightarrow V\uparrow$
$\Rightarrow V>>b$
$\therefore$ Neglect $b$
Van der Waal's equation $=(P+\frac{a}{V^2})V=RT$
At high pressure,
$P$ is very large
$\Rightarrow P>>\frac{a}{V^2}$
$\therefore$ Neglect $\frac{a}{V^2}$
Van der Waal's equation $=P(V-b)=RT$
At low pressure and high temperature, real gas behaves as an ideal gas. So, Van der Waal's equation reduces to
$PV=RT$
Hence, option (b) is correct.