First Law of Thermodynamics Ideal gas law In a constant pressure process, the change in internal energy is 
$$
U = \frac{3}{2} nRT = \frac{3}{2} P(V2 - V1).
$$
However in a constant volume, is the change in internal energy $U = \frac{3}{2} nRT = V(P2-P1)$?
 A: The thermodynamics of an ideal gas is entirely contained in two equations:
$$
U = \frac{3}{2}nRT ~~~~~~~~~~~~~~~~~~~~~~~[1]\\
PV=nRT.~~~~~~~~~~~~~~~~~~~~~~~[2]
$$
In a transformation (change of state) where pressure is kept fixed, equation [2] says that if the volume changes from $V_1$ to $V_2$ the temperature cannot remain fixed at the same value T. The change of volume must be accompanied by a change of temperature $\Delta T = P \Delta V/nR$.
In a similar way, if volume is kept fixed, again from eq.[2] we can get the change of temperature corresponding to a change of pressure from $P_1$ to $P_2$.
From the change of temperature, eq.[1] immediately provides the change of internal energy.
A: Change in Internal Energy is given by
$$\triangle U =nC_{v}\triangle T\ $$
$$\triangle U =n\left(\frac{fR}{2}\right)\triangle T\ $$
Assuming f=3 in your case,
$$\triangle U =n\left(\frac{3R}{2}\right)\triangle T\  \;\;\;\;\;\;[1] $$
Now, ideal gas equation says,
$$PV=nRT\;\;\;\;\;\;[2]$$
At constant pressure,
$$\ nR\triangle T=P \triangle V  \;\;\;\;\;\;[3]$$
From equation [3] in equation [1] ,
$$\triangle U =n\left(\frac{3R}{2}\right)P \triangle V \  \;\;\;\;\;\; $$
At constant volume ,
$$\ nR\triangle T=V\triangle P  \;\;\;\;\;\;[4]$$
From equation [4] in equation [1] ,
$$\triangle U =n\left(\frac{3R}{2}\right)V \triangle P \  \;\;\;\;\;\; $$
