Question

1. 0.1 m3 of an ideal gas at 300 K and 1 bar is compressed adiabatically to 8

bar. It is then cooled at constant volume and further expanded isothermally so as to reach the

condition from where it started. Calculate :

(i) Pressure at the end of constant volume cooling.

(ii) Change in internal energy during constant volume process.

(iii) Net work done and heat transferred during the cycle. Assume

cp = 14.3 kJ/kg K and cv = 10.2 kJ/kg K.

Answer 1

$V_1 = 0.1 \;m^3 \\ T_1 = 300 \;K \\ p_1 = 1\; bar \\ c_p = 14.3 \;kJ/kg\; K \\ c_v = 10.2 \;kJ/kg \;K$

(i) Pressure at the end of constant volume cooling, p₃ :

$γ=\frac{c_p}{c_v} = \frac{14.3}{10.2}=1.402$

Characteristic gas constant,

$R=c_p -c_v = 14.3 -10.2 = 4.1 \;kJ/kg \; K$

Considering process 1-2, we have :

$p_1V_1^γ =p_2V_2^γ \\ V_2 = V_1 \times (\frac{p_1}{p_2})^{1/γ} \\ = 0.1 \times (\frac{1}{8})^{1/1.402} \\ = 0.0227 \;m^3 \\ \frac{T_2}{T_1} = (\frac{p_2}{p_1})^{(γ-1)/γ} \\ = (\frac{8}{1})^{(1.402-1)/1.402} \\ = 1.815 \\ T_2 =T_1 \times 1.815 \\ = 300 \times 1.815 \\ = 544.5 \;K$

Considering process 3–1, we have

$p_3V_3=p_1V_1 \\ p_3 = \frac{p_1V_1}{V_3} \\ = \frac{1 \times 0.1}{0.0227} \\ = 4.4 \;bar$

(ii) Change in internal energy during constant volume process, (U₃ – U₂) :

Mass of gas,

$m=\frac{p_1V_1}{RT_1} \\ = \frac{1 \times 10^5 \times 0.1}{4.1 \times 1000 \times 300} \\ = 0.00813 \;kg$

Change in internal energy during constant volume process 2–3,

$U_3 -U_2 =mc_v(T_3 -T_2) \\ = 0.00813 \times 10.2(300-544.5) \\ = -20.27 \;kJ \\ T_3 =T_1$

(– ve sign means decrease in internal energy)

During constant volume cooling process, temperature and hence internal energy is reduced. This decrease in internal energy equals to heat flow to surroundings since work done is zero.

(iii) Net work done and heat transferred during the cycle :

$W_{1-2} = \frac{p_1V_1 -p_2V_2}{γ-1} = \frac{mR(T_1-T_2)}{γ-1} \\ = \frac{0.00813 \times 4.1(300-544.5)}{1.402-1} \\ = -20.27 \;kJ \\ W_{2-3}=0$

... since volume remains constant

$W_{3-1} = p_3V_3 log_e(\frac{V_1}{V_3}) \\ = p_1V_1 log_e (\frac{p_3}{p_1}) \\ = 1 \times 10^5 \times 0.1 \times log_e(\frac{4.4}{1}) \\ = 14816 \;Nm \\ = 14.82 \;J$

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

$= (-20.27) + 0 + 14.82 = -5.45 \;kJ$

–ve sign indicates that work has been done on the system.

For a cyclic process :

$\oint δQ= \oint δW$

Heat transferred during the complete cycle = – 5.45 kJ

–ve sign means heat has been rejected i.e., lost from the system.