Question

One mole of an ideal gas undergoes a process $P=P_0{\left(1+{\left(\frac{2V_0}{V}\right)}^2\right]}^{-1},$ where$P_0,V_0$ are constants. Change in temperature of the gas when volume is changed from $V=V_0$ to $V=2V_0$ is

• A. $\frac{4}{5}\frac{P_0{V}_0}{nR}$
• B. $\frac{3}{4}\frac{P_0{V}_0}{nR}$
• C. $\frac{2}{3}\frac{P_0{V}_0}{nR}$
• D. $\frac{9}{7}\frac{P_0{V}_0}{nR}$

Answer 1

The correct option is A $\frac{4}{5}\frac{P_0{V}_0}{nR}$

$P=P_0{\left(1+{\left(\frac{2V_0}{V}\right)}^2\right]}^{-1}AtV=V_0,P=\frac{P_0}{5}{T}_i=\frac{PV}{nR}=\frac{\left(\frac{P_0}{5}\right)V_0}{nR}=\frac{P_0{V}_0}{5nR}AtV=2V_0,P=\frac{P_0}{2}{T}_f=\frac{PV}{nR}=\frac{\left(\frac{P_0}{2}\right)2V_0}{nR}=\frac{P_0{V}_0}{nR}\Delta T=T_f-T_i=\frac{4P_0{V}_0}{5nR}$