Question

Ammonia weighing 22 kgs is confined inside a cylinder equipped with a piston has an initial pressure of 413 kPa at 38°C. If 2900 kJ of heat is added to the ammonia until its final pressure and temperature are 413 kPa at 100°C, respectively, what is the amount of work done by the fluid in kJ?                                                                             

Answer 1

$P_1=P_2=413 \;kPa \\ T_1 = 38 \;°C \\ T_2 = 100 \;°C \\ ΔT=100-38=62 \;°C \\ m_1 = 22 \;kg \\ Q = 2900 \;kJ$

Work formula

$W_{nf} = P(V_2-V_1) \\ = mR(T_2 -T_1) \\ R = \frac{\bar{R}}{MW} = \frac{8.3145}{14 + 3(1.298)} = 0.464 \;kJ/kg \cdot K \\ W_{nf} = 22 \times 0.464 \times 62 = 633.786 \;kJ$

Answer: 633.786 kJ