Question

One mole of an ideal gas u ndergoes a process:
$P=\frac{P_0}{1+(V_0/V{)}^2}$.
Here $P_0$ and $V_0$ are constants. change in temperature of the gas when volume is changed from $V=V_0$ to $V=2V_0$ is:

• A. $-\frac{2P_0{V}_0}{5R}$
• B. $\frac{11P_0{V}_0}{10R}$
• C. $-\frac{5P_0{V}_0}{4R}$
• D. $P_0{V}_0$

Answer 1

The correct option is B $\frac{11P_0{V}_0}{10R}$

$Given,P=\frac{P_0}{1+{\left(\frac{V_0}{V}\right)}^2}Numberofmolesofgas,n=1For{V}_1=V_0,P_1=\frac{P_0}{1+1^2}=\frac{P_0}{2}For{V}_2=2V_0,P_2=\frac{P_0}{1+(\frac{1}{2}{)}^2}=\frac{4}{5}{P}_{0}Usingidealgasequation,PV=nRT{P}_1{V}_1=R{T}_1{T}_1=\frac{P_0{V}_0}{2R}{P}_2{V}_2=R{T}_2{T}_2=\frac{8P_0{V}_0}{5R}\Delta T=T_2-T_1=(\frac{8}{5}-\frac{1}{2})\frac{P_0{V}_0}{R}=\frac{11P_0{V}_0}{10R}Hence,option\left(B\right)iscorrect.$