Answer to Question #190743 in Mechanical Engineering for Isaac Ferkah

Question

Calculate the change in internal energy, heat transfer and change in enthalpy for 0.5 kg of air expanding according to law PV1.2 = C from 1 MPa and 300⁰C to 100 kPa. What will be the work done by air during expansion?   [ Cv = 0.718 kJ/kg K for air] 

Answer 1

$\Delta U=c_v m (T_2-T_1)$

$T_2=T_1\frac{p_2}{p_1}\left(\frac{p_1}{p_2}\right)^\frac{1}{1.2}\\ T_2=(300+273)\frac{1}{10}\left(10\right)^\frac{1}{1.2}=390\ K$

$\Delta U=(0.718)(0.5) (390-573)=-65.7\ kJ$

$Q=\frac{8.31}{29}(0.5) (390-573)=-26.2\ kJ$

$W=(-26.2)\frac{1.395-1.2}{1.395-1}=-12.9\ kJ$