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InterviewSolution
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Two moles of a certain gas at a temperature `T_0=300K` were cooled isochorically so that the pressure of the gas got reduced 2 times. Then as a result of isobaric process, the gas is allowed to expand till its temperature got back to the initial value. Find the total amount of heat absorbed by gas in this process. |
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Answer» Let `v = 2` moles of the gas. In the first phase, uder isochoric process, `A_1 = 0`, therefore from gas law if pressure is reduced `n` times so that temperature i.e., new temperature becomes `T_0//n`. Now from first law of thermodynamics `Q_1 = Delta U_1 = (v R Delta T)/(gamma - 1)` =`(v R)/(gamma - 1)((T_o)/(n) -T_0) = (v RT_0 (1 - n))/(n(gamma - 1))` During the second phase (under isobaric process), `A_2 = p Delta V = v R Delta T` Thus from law of thermodynamics : `Q_2 = Delta U_2 + A_2 = (v R Delta T)/(gamma - 1)+ v R Delta T` =`(v R(T_0 - (T_0)/(n))gamma)/(gamma - 1) = (vRT_0(n - 1)gamma)/(n(gamma - 1))` Hence the total amount of heat absorbed `A = Q_1 + Q_2 = (v RT_0(1 -n))/(n(gamma - 1))+(v RT_0(n - 1)gamma)/(n(gamma - 1))` =`(vRT_0(n - 1)gamma)/(n(gamma - 1))(-1 + gamma) = V RT_0(1 - (1)/(n))`. |
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