An ideal gas undergoes a transition from state 1 to state 2 and then back to sta
ID: 2002744 • Letter: A
Question
An ideal gas undergoes a transition from state 1 to state 2 and then back to state 1, along the closed path shown in the figure. Let ?U stand for the change of the internal energy of the gas over the full cycle 1?2?1, W be the work done on the gas by its environment and Q be the heat added to the gas over the full cycle.
What can be said about the values ?U, W and Q?
?U = 0, W > 0, Q > 0
?U = 0, W < 0, Q > 0
?U < 0, W > 0, Q < 0
?U > 0, W = 0, Q < 0
?U = 0, W > 0, Q > 0
?U = 0, W < 0, Q > 0
?U < 0, W > 0, Q < 0
?U > 0, W = 0, Q < 0
Pres Pressure VolumeExplanation / Answer
Ans: The correct physical processes is U = 0, W < 0, Q > 0 ;
Reasons:(i) Since the gas goes through a cyclic path it brought back to its initial state and since internal energy is a function of state only so U = 0;
(ii) The area under the PV curve shows the amount of work done by the gas.Here the gas through 12 and brought back through 21. now the area under 12 curve is greater than the area under 21 curve. So the amount of work done by the gas is positive ,Hence the work done by the external egent is negative, so, W<0.
(ii) The amount of heat added to the gas will be positive to satisfy the energy conservation. And it directly follows from 1st law of thermodynamics dQ=dU+ dW.
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