Question

One mole of an ideal gas passes through a process where pressure and volume obey the relation $P=P_o\left[1-\frac{1}{2}{\left(\frac{V_0}{V}\right)}^2\right]$ .Here $P_o$ and $V_o$ are constants. Calculate the change in the temperature of the gas if its volume changes from $V_o$ to $2V_o$ .

• A. $\frac{1}{2}\frac{P_o{V}_o}{R}$
• B. $\frac{3}{4}\frac{P_o{V}_o}{R}$
• C. $\frac{5}{4}\frac{P_o{V}_o}{R}$
• D. $\frac{1}{4}\frac{P_o{V}_o}{R}$

Answer 1

Answer: (C)

$P=P_0\left[1-\frac{1}{2}{\left(\frac{V_0}{V}\right)}^2\right]$
Pressure at $V_0=P_0\left(1-\frac{1}{2}\right)=\frac{P_0}{2}$
Pressure at $2V_0=P_0\left(1-\frac{1}{2}\times \frac{1}{4}\right)=\frac{7}{8}{P}_0$
Temperature at $V_0=\frac{\frac{P_0}{2}{V}_0}{nR}=\frac{P_0{V}_0}{2nR}$
Temperature at $2V_0=\frac{\left(\frac{7}{8}{P}_0\right)\left(2V_0\right)}{nR}=\frac{\frac{7}{4}{P}_0{V}_0}{nR}$
$=\left(\frac{7}{4}-\frac{1}{2}\right)\frac{P_0{V}_0}{nR}$
Change in temperature$=\frac{5}{4}\frac{P_0{V}_0}{nR}=\frac{5P_0{V}_0}{4R}$