Question

. 0.15 m3 of an ideal gas at a pressure of 15 bar and 550 K is expanded
isothermally to 4 times the initial volume. It is then cooled to 290 K at constant volume and then
compressed back polytropically to its initial state.
Calculate the net work done and heat transferred during the cycle.

Answer 1

Given :

$V_1 = 0.15 \;m^3 \\ p_1 = 15 \; bar \\ T_1 = T_2 = 550\; K \\ \frac{ V_2}{V_1} = 4 \\ T_3 = 290 \; K$

Considering the isothermal process 1–2, we have

$p_1 V_1 = p_2 V_2 \\ p_2 = \frac{p_1V_1}{V_2} \\ p_2 = \frac{15 \times 0.15}{4 \times 0.15} = 3.75 \;bar$

Work done,

$W_{1-2} = p_1V_1 log_e(\frac{V_2}{V_1}) \\ = (15 \times 10^5 ) \times 0.15 \times log_e (4) \\ = 311916 \; J = 311.9 \; kJ$

Considering constant volume process 2-3, we get

$V_2=V_3=V_4 \times 0.15 =0.6 \; m^3 \\ \frac{p_2}{T_2} = \frac{p_3}{T_3} \\ p_3=p_2 \times \frac{T_3}{T_2} = 3.75 \times \frac{290}{550} = 1.98 \;bar \\ W_{2-3} =0$

...since volume remains constant

Consider polytropic process 3–1 :

$p_3V_3^n = p_1V_1^n \\ \frac{p_1}{p_3} = (\frac{V_3}{V_1})^n$

Taking log on both sides, we get

$log_e(\frac{p_1}{p_3}) = n log_e( \frac{V_3}{V_1}) \\ n = \frac{log_e(\frac{p_1}{p_3})}{log_e( \frac{V_3}{V_1})} \\ = \frac{log_e(\frac{15}{1.98})}{log_e(4)} \\ = 1.46 \\ W_{3-1} = \frac{p_3V_3 -p_1V_1}{n-1} \\ = \frac{1.98 \times 10^5 \times 0.6 – 15 \times 10^5 \times 0.15}{1.46-1} \\ = -230869 \;J \\ = -230.87 \;kJ$

Net work done $= W_{1–2} + W_{2–3} + W_{3–1}$

$= 311.9 + 0 +(-230.87) = 81.03 \;kJ$

For a cuclic process,

$\oint δQ= \oint δW$

Heat transferred during the cycle = 81.03 kJ