InterviewSolution
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InterviewSolution
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Figure shows a process `ABCA` performed on an ideal gas. Find the net heat given to the system during the process. ` |
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Answer» Since the process is cyclic , hence the change in internal energy is zero. The heat given to the system is then equal to the work done by it. The work done in part `AB` is `W_(1) = 0` (the volume remains constant ) . The part `BC` represents an isothermal process so that the work done by the gas during this part is `W_(2) = nRT_(2) "ln"(V_(2))/(V_(1))` During the part `CA: Vprop T " "` So,`V//T` is constant and hence , `P=(nRT)/(V)` is constant The work done by the gas during the part `CA "is" W_(3) = P(V_(1)-V_(2)) = nRT_(1) -nRT_(2) = -nR(T_(2)-T_(1))`. The net work done by the gas in the process `ABCA` is `W= W_(1) + W_(2) + W_(3) = "n"R[T_(2)"ln"(V_(2))/(V_(1)) - (T_(2)-T_(1))]` The same amount of heat is given to the gas. |
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