Question

For one mole of ideal gas if $p=\frac{P_0}{1+(\frac{V}{V_0}{)}^2}$ where $P_0$ & $V_0$ are constant then temperature of gas when $V=V_0$ is:

• A. $\frac{P_0{V}_0}{R}$
• B. $\frac{P_0{V}_0}{4R}$
• C. $\frac{P_0{V}_0}{2R}$
• D. $\frac{P_0{V}_0}{3R}$

Answer 1

The correct option is C $\frac{P_0{V}_0}{2R}$

Solution:- (C) $\frac{P_{\circ }{V}_{\circ }}{2R}$

$P=\frac{P_{\circ }}{1+{\left(\frac{V}{V_{\circ }}\right)}^2}$

$\because V=V_{\circ }\left(\text{Given}\right)$

$P=\frac{P_{\circ }}{1+{\left(\frac{V_{\circ }}{V_{\circ }}\right)}^2}$

$\Rightarrow P=\frac{P_{\circ }}{1+1}=\frac{P_{\circ }}{2}$

From ideal gas equation,

$PV=nRT$

$\Rightarrow T=\frac{PV}{nR}$

$\because n=1\left(\text{Given}\right)$

$\therefore T=\frac{\left(\frac{P_{\circ }}{2}\right)\cdot V_{\circ }}{1\times R}=\frac{P_{\circ }{V}_{\circ }}{2R}$