One mole of an ideal gas passes through a process where pressure and volume obey the relation P = P$$_0$$[1 - $$\frac{1}{2}\left(\frac{V_0}{V}\right)^2$$]. Here P$$_0$$ and V$$_0$$ are constants. Calculate the change in the temperature of the gas if its volume changes from V$$_0$$ to 2V$$_0$$.

For one mole of an ideal gas we always have the ideal-gas equation

$$PV = nRT\,,$$

and for one mole $$n = 1$$, so it reduces to

$$PV = RT \;. \quad -(1)$$

The problem gives the pressure-volume relation

$$P = P_0\!\left[\,1 - \dfrac12\left(\dfrac{V_0}{V}\right)^2\right]. \quad -(2)$$

The gas volume changes from the initial value $$V_1 = V_0$$ to the final value $$V_2 = 2V_0$$. We have to find the corresponding temperatures and then their difference.

First we examine the initial state. Putting $$V = V_0$$ in equation (2) we get

$$P_1 = P_0\!\left[\,1 - \dfrac12\left(\dfrac{V_0}{V_0}\right)^2\right] = P_0\!\left[\,1 - \dfrac12(1)^2\right] = P_0\!\left(1 - \dfrac12\right) = \dfrac{P_0}{2}\;.$$ \quad -(3)

Using equation (1) to obtain the initial temperature:

$$T_1 = \dfrac{P_1 V_1}{R} = \dfrac{\left(\dfrac{P_0}{2}\right)V_0}{R} = \dfrac{P_0 V_0}{2R}\;. \quad -(4)$$

Now we analyse the final state. Put $$V = 2V_0$$ in equation (2):

$$P_2 = P_0\!\left[\,1 - \dfrac12\left(\dfrac{V_0}{2V_0}\right)^2\right] = P_0\!\left[\,1 - \dfrac12\left(\dfrac12\right)^2\right] = P_0\!\left[\,1 - \dfrac12\left(\dfrac14\right)\right] = P_0\!\left[\,1 - \dfrac18\right] = P_0\!\left(\dfrac78\right) = \dfrac{7P_0}{8}\;. \quad -(5)$$

Again applying equation (1) for the final temperature:

$$T_2 = \dfrac{P_2 V_2}{R} = \dfrac{\left(\dfrac{7P_0}{8}\right)(2V_0)}{R} = \dfrac{7P_0 V_0}{4R}\;. \quad -(6)$$

The change in temperature is

$$\Delta T = T_2 - T_1 = \dfrac{7P_0 V_0}{4R} - \dfrac{P_0 V_0}{2R}.$$

To combine the fractions we express the second term with denominator $$4R$$:

$$\dfrac{P_0 V_0}{2R} = \dfrac{2P_0 V_0}{4R}.$$

So

$$\Delta T = \dfrac{7P_0 V_0}{4R} - \dfrac{2P_0 V_0}{4R} = \dfrac{5P_0 V_0}{4R}\;. \quad -(7)$$

Hence, the correct answer is Option B.