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Asked by sarveshvibrantacademy | 12 May, 2019, 01:47: PM
answered-by-expert 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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