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Asked by sarveshvibrantacademy | 12 May, 2019, 01:47: PM
Expert Answer
(A) internal energy decreases in the path 3-4
True.
By first law of thermodynamics, dq = du + dw ....................(1)
where dq is the heat energy input , du is change in internal energy and dw is the workdone by the gas.
for adiabatic process, dq = 0, workdone by the gas dw= ∫pdv is positive,
hence du = -dw or internal energy is decreasing.
(B) no work is done during the path 2-3, because volume is constant and change in volume dv = 0
(C) during the path 4-1, volume is constnat.
For ideal gas when volume is constant pressure P is directly proportional to Temperature T.
Since pressure decreasing in the path 4-1, Temperature also devreasing.
(D) during the path 1-2 , workdone dw= ∫pdv , change of volume is negative.
Hence workdone by the gas is negative. Hence work is done on the gas
Answered by Thiyagarajan K | 12 May, 2019, 03:26: PM
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