Question 4.14: Problem: An ideal gas with cp = 1.044 kJ / kgK and cv = 0.74...
Problem: An ideal gas with c_p = 1.044 kJ / kgK and c_v = 0.745 kJ / kgK contained in a cylinder–piston assembly initially has a pressure of 150 kPa, a temperature of 30 °C, and a volume of 0.22 m³ . It is heated slowly at constant volume (process 1–2) until the pressure is doubled. It is then expanded slowly at constant pressure (process 2–3) until the volume is doubled. Determine the work done and heat added in the combined process.
Find: Work done W and heat added Q.
Known: Ideal gas, specific heats c_p = 1.044 kJ / kgK and c_v = 0.745 kJ / kgK, initial volume V_1 = 0.22 m³ , initial pressure P_1 = 150 kPa, initial temperature T_1 = 30 °C, final pressure P_2 = 2 P_1 , final volume V_3 = 2 V_1 .
Diagram:
Assumptions: No change in KE and PE of system, constant specific heats.
Governing Equations:
Ideal gas equation PV = mRT
First Law Q_{12}+ W_{12}=\Delta U_{12}
The gas constant for this ideal gas is R = c_ p – \ c_v = 1.044 \ kJ / kgK – \ 0.745 \ kJ / kgK = 0.299 \ kJ / kgK.
To find the mass of gas in the cylinder,
m=\frac{P_1V_1}{RT_1} =\frac{150 \times 10^3 \ N/m^2 \times 0.22 \ m^3}{0.299 \ kJ/kgK \times (273.15 + 30) \ K} =0.364 \ 070 \ kg.
The temperature at state 2 is
T_2 = T_1\frac{P_2}{P_1} =303.15 \ K \times 2 =606.300 \ K.
The temperature at state 3 is
T_3=T_2\frac{V_3}{V_2} =606.3 \ K \times 2 =1212.60 \ K.
Work done during process 1–2 is
W_{12}=-\int\limits_{V_1}^{V_2}{P}dV =0,
with heat transfer during process 1–2 of
Q_{12}=\Delta U_{12}-{W}/{_{12}}=\Delta U_{12}=mc_{v}(T_2-T_1).
Q_{12}=0.364 \ 07 \ kg \times 0.745 \ kJ/kgK (606.3 \ K-303.15 \ K)=82.2240 \ kJ.
Work done during process 2–3 is
W_{23}=-\int\limits_{V_2}^{V_3}{P} dV=-P_3(V_3-V_2)=-(P_3V_3-P_2V_2)=-mR(T_3-T_2),
W_{23}=-0.364 \ 07 \ kg \times 0.299 \ kJ/kgK (1212.6 \ K -606.3 \ K)=-66.0000 \ kJ.
Heat transfer during process 2–3 is
Q_{23}=\Delta U_{23}-W_{23}=mc_v(T_3-T_2)-W_{23},
Q_{23}=0.364 \ 07 \ kg \times 0.745 \ kJ/kgK(1212.6 \ K-606.3 \ K)+66 \ kJ=230.448 \ kJ.
Total work done is
W_{total}=W_{12}+W_{23}=-66 \ kJ.
Total heat transfer is
Q_{total}=Q_{12}+Q_{23}=82.224 \ kJ+230.448 \ kJ=312.672 \ kJ.
Answer: The system does 66.00 kJ of work on the surroundings while 312.7 kJ of heat is added to it.