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
Expert's answer

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




"\\frac{p_1}{T_1}=\\frac{p_2}{T_2},\\\\\n\\space\\\\\nT_2=T_1\\frac{p_2}{p_1}=2763\\text{ K}."


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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