Answer to Question #98631 in Electrical Engineering for Hardi Amiru Sulemana

Question

a quantity of a gas occupies 0.14 m cube at 9.65 bar and 371 degrees Celsius is heated during a constant volume process and the pressure reaches 41.4 bar. The gas is then expanded adiabatically to a pressure of 2.76 bar. Given that R is equal to 0.288 kilo joules per kilogram Kelvin and Gamma is equal to 1.41, calculate:
a) the temperature at the beginning of the expansion
b) the temperature at the end of the expansion
c) the heat energy supplied
d) the work energy transferred

Answer 1

a) The temperature at the beginning can be found from the condition

$$\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2},$$

where

$$V_1=V_2:$$

$$\frac{p_1}{T_1}=\frac{p_2}{T_2},\\ \space\\ T_2=T_1\frac{p_2}{p_1}=2763\text{ K}.$$

Hot!

b) The adiabatic law (and keep in mind that for isochoric process $V_1=V_2$):

$$p_2V_1^\gamma=p_3V_3^\gamma=\text{ const}=c,\\ \space\\ V_3=V_1\bigg(\frac{p_2}{p_3}\bigg)^{1/\gamma}=0.14\bigg(\frac{41.4}{2.76}\bigg)^{(1/1.41)}=0.95\text{ m}^3.$$

c) The word adiabatic means that no energy of heat is lost or supplied.

d) The work done in the adiabatic process while gas transferred from state 2 (0.14 cubic m, 41.4 bar) to state 3 (2.76 bar, 0.95 cubic m):

$$W=\int^3_2pdV=\int^3_2\frac{c}{V^{1.41}}dV=-\frac{c}{0.41V^{0.41}}\bigg|^3_2=\\ \space\\ =-\frac{c}{0.41V_3^{0.41}}+\frac{c}{0.41V_2^{0.41}}= \space\\ \space\\ =-\frac{p_3V_3^{1.41}}{0.41V_3^{0.41}}+\frac{p_2V_2^{1.41}}{0.41V_2^{0.41}}=\frac{p_2V_2-p_3V_3}{0.41}=774146\text{ J}.$$