Question

A fluid contained in a cylinder receives 150 kJ of mechanical energy by means of a paddle wheel, together with 50 kJ in the form of heat. At the same time, a piston in the cylinder moves in such a way that the pressure remains constant at 200 kN/m${}^2$ during the fluid expansion from 2 m${}^3$ to 5m${}^3$. The change in enthalpy is

• A. 200 kJ
• B. - 400 kJ
• C. - 200 kJ
• D. 400 kJ

Answer 1

The correct option is A 200 kJ

$\text{Given:}P=200kN/m^2;$

$W_1=-150kJ;$ $Q=50kJ$

$\text{We know,}H=U+PV$

$H_1=U_1+P_1{V}_1$

$H_2=U_2+P_2{V}_2$

$\therefore \text{Change in enthalpy}$

$\Delta H=H_2-H_1=(U_2-U_1)+P(V_2-V_1)$

Work done by the system,

$W_2=P(V_2-V_1)=200\times (5-2)=600kJ$

From 1st law of thermodynamics,

$\Delta Q=\Delta U+\Delta W$

$\Delta U=\Delta Q-\Delta W=50-(-150+600)=50-450=-400kJ$

So, $\Delta H=\Delta U+P(V_2-V_1)$

$\therefore$ $\Delta H=-400+200(5-2)=-400+600=200kJ$