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Asked by kkudgiri55 17th October 2018, 6:13 PM
For monoatomic ideal gas, Heat absorbed/released at constant volume process = n×CV×ΔT = n×(3/2)×R×ΔT..................(1)
where n is number of moles, CV is specific heat at constant volume, R is gas constant and ΔT is increase/decreas of temperature

For monoatomic ideal gas, Heat absorbed/released at constant pressure process = n×CP×ΔT = n×(5/2)×R×ΔT..................(2)

Let us start from point D in the given P-V diagram.
Let us assume when the gas is at the state represented by point D in the P-V diagram, its temperature is T0 .

When the gas pressure increases from P0 to 2P0 at constant volume V0 during the process represented by DA in the P-V diagram,
its temperaure is increased from T0 to 2T0 .

Hence heat absorbed in the DA part of cycle  is, QDA = (3/2)×n×R×T0 ................................................(3)

For the AB part of cycle, Volume doubles at constant pressure. Hence temperature at B is 4T0 .
Heat aborbed in the AB part of cycle, QAB = (5/2)×n×R×2T0 = 5×n×R×T0 .................................................(4)

For the BC part of cycle, Pressure is decreased to half. Hence temperature at C is 2T0 .
Heat released in the BC part of cycle, QBC = -(3/2)×n×R×2T0 = -3×n×R×T0 .................................................(5)

Heat released in the CD part of cycle, QCD = -(5/2)×n×R×T .................................................(6)

net Amount of heat absorbed in full cycle = QDA + QAB + QBC + QCD = [ (3/2) + 5 - 3 - (5/2) ]×n×R×T0 = n×R×T0 = P0×V0

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Answer to this problem can be obtained in alternate way.

For a complete cycle, at a given point in P-V diagram that represents the state of the gas, change in internal energy ΔE is zero.

Hence ΔE = ΔQ+ΔW =0  ..............................(7)

we have ΔW = workdone in the path AB + workdone in the path CD = -(2P0×V0 ) + ( P0×V0) = - ( P0×V0) ...................(8)

from eqn.(7) and (8), we have ΔQ = - ΔW = P0×V0
Answered by Expert 17th October 2018, 10:50 PM
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Tags: heat-energy

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