An ideal gas undergoes the process A=>B=>C=>A. The heat gained in process AB is

Question

An ideal gas undergoes the process A=>B=>C=>A. The heat gained in process AB is 576 J, while in process BC the system loses 67 J. In process AB the system performs 320 J of work, while in process CA work is done on the system in the amount of 238 J. How much heat is transferred to or from the system in process CA?

Explanation / Answer

1. A. - 427 J Explanation: The gas undergoes a cycle process. Since internal energy is state function, i.e. each state has a certain internal energy, the total change in internal energy for such a process, in which you return to some initial state is zero. Furthermore the change in internal energy equals net heat added to the gas in the cycle minus net work done by the gas. Thus heat and work have the same magnitude. Symbolically: ?U = Q - W = 0 => Q = W The net work done is the sum of the works done in each step. W = W_AB + W_BC + W_CA Work for both AB and CA is given. The work done in step BC is zero, because the volume is constant in this part of the process. Hence W = 320 J + 0 J - 238 J = 82 J (Note that W_BC is negative because work is done ON the gas) Similarly you can compute the net heat added to the gas: Q = Q_AB + Q_BC + Q_CA So heat added in step CA is: Q_CA = Q - Q_AB - Q_BC with Q = W Q_CA = W - Q_AB - Q_BC = 82 J - 576 J - (-67J) = -427 J

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