Solution:

For one mole of an ideal gas, we use the ideal gas law:

\[
PV = RT
\]

Given \( P = \frac{P_0}{1 + \left( \frac{V_0}{V} \right)^2} \), find \( T \) at \( V = V_0 \) and \( V = 2V_0 \).

1. When \( V = V_0 \):
\[
P = \frac{P_0}{1 + \left( \frac{V_0}{V_0} \right)^2} = \frac{P_0}{1 + 1} = \frac{P_0}{2}
\]
\[
T_1 = \frac{PV}{R} = \frac{\frac{P_0}{2} \cdot V_0}{R} = \frac{P_0 V_0}{2R}
\]

2. When \( V = 2V_0 \):
\[
P = \frac{P_0}{1 + \left( \frac{V_0}{2V_0} \right)^2} = \frac{P_0}{1 + \frac{1}{4}} = \frac{P_0}{\frac{5}{4}} = \frac{4P_0}{5}
\]
\[
T_2 = \frac{PV}{R} = \frac{\frac{4P_0}{5} \cdot 2V_0}{R} = \frac{8P_0 V_0}{5R}
\]

3. Change in temperature:
\[
\Delta T = T_2 - T_1 = \frac{8P_0 V_0}{5R} - \frac{P_0 V_0}{2R} = \frac{11P_0 V_0}{10R}
\]