Consider one mole of helium at temperature 140K, that suffers a free expansion of Joule between 1L and 2L. Admit that the specific heat at constant value is Cv=(3/2)R. Using the general expression of the variation of internal energy $$dU = n C_V dT + [T(\frac{dp}{dt})-p]dV$$ where $$p=\frac{nRT}{V-nb}-\frac{n^2a}{V^2}$$ $$a=3.44 \times 10^{-3} Pa m^6 mol^{-2}$$ $$b= 23.40 \times 10^{-6} m^3 mol^{-1}$$

My attempt: I first calculated $$\frac{dp}{dt}= \frac{nR}{V-nb}$$

Substituting in the equation of energy $$dU = n C_V dT + [\frac{nRT}{V-nb}-p]dV$$

Integrating $$U = n C_V T + [nRT\ln{V-nb}-pV]$$

Now in a free expansion of Joule the variation of U is zero. Then: $$n C_V T_1 + [nRT_1\ln{V_1-nb}-p_1V_1]= n C_V T_2 + [nRT_2\ln{V_2-nb}-p_2V_2]$$

Solving for T2, we obtain

$$T_2=\frac{n3/2RT_1+nRT_1\ln(V_1-nb)-\frac{nRT_1V_1}{V_1-nb}+\frac{n^2a}{V_1}-\frac{n^2a}{V_2}}{n3/2R+nR\ln(V_2-nb)-\frac{nRV_2}{V_2-nb}}$$

Substituting all the values I obtain 157.5 K. However the answer should 139.9. Am I missing something?