Answer
zero
Work Step by Step
For state 3, the ideal gas equation can be written as:
$P_3V_3=nRT_3$
This equation can be rearranged as:
$T_3=\frac{P_3V_3}{nR}$...............eq(1)
Furthermore,
$P_3=2$
$P_1=2(20.2\times10^5)=40.4\times10^5Pa$
and $V_3=0.500V_1=0.500(0.0015)=0.00075m^3$
Putting these values in eq(1), we get
$T_3=\frac{(40.4\times10^5)(0.00075)}{(3)(8.314)}$
$T_3=122K$
As $T_3=T_1$ so $\Delta T=0$ This shows that change in internal energy is zero from state 1 to state 3.
