InterviewSolution
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InterviewSolution
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`2m^(3)` volume of a gas at a pressure of `4xx10^(5) Nm^(-2)` is compressed adiabatically so that its volume becomes `0.5m^(3)` Find the new pressure . Compare this with the pressure that would result if the compression was isothermal. Calculate work done in each `gamma = 1.4` |
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Answer» `V_(1) = 2m^(3), P_(1) = 4xx10^(5)Nm^(-2)` , `" "` `V_(2) = 0.5m^(3)` In adiabatic process `" "` `P_(1)V_(1)^(gamma) = P_(2)V_(2)^(gamma) implies P_(2) = 4xx10^(5)[(2)/(0.5)]^(1.4) = 4xx10^(5) (4)^(1.4) = 2.8xx 10Nm^(-2)` In isothermal process `P_(1)V_(1) = P_(2)V_(2) implies P_(2) = (P_(1)V_(1))/(V_(2)) = (4xx10^(5)xx2)/(0.5) = 1.6 xx 10^(6)Nm^(-2).` Now work done in adiabatic process `W = (P_(1)V_(1) - P_(2)V_(2))/(gamma-1) = ((4xx10^(5) xx 2 -2.8 xx 10^(6) xx 0.5))/(1.4-1) = -1.48 xx10^(6)J` "Work done in isothermal process" `W= 2.3026RT log(V_(2))/(V_(1)) = 2.3026P_(1)V_(1)log(V_(2))/(V_(1))` ` = 2.3026 xx 4 xx 10^(5) xx 2 xx log[(0.5)/(2.0)] = 2.3026 xx 4xx 10^(5) xx 2log((1)/(4)) = -1.1 xx 10^(6)J` |
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