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

Question #98631
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
1
2019-11-15T07:07:38-0500

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},\\\\\n\\space\\\\\nT_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,\\\\\n\\space\\\\\nV_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=\\\\\n\\space\\\\\n=-\\frac{c}{0.41V_3^{0.41}}+\\frac{c}{0.41V_2^{0.41}}=\n\\space\\\\\n\\space\\\\\n=-\\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}."

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