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Question

One mole of an ideal gas is taken from state A to state B by three different processes (a) ACB, (b) ADB and (c) AEB as shown in the PV diagram. The heat absorbed by the gas is


A
Greater in process (b) than in (a)
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B
The least in process (b)
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C
The same in (a) and (c)
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D
Less in (c) than in (b)
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Solution

The correct option is D Less in (c) than in (b)
From first law of thermodynamics Heat absorbed by gas in three processes is given by,
QACB=ΔU+WACB ...(i)
QADB=ΔU ...(ii)
(WADB=0, as V= constant)
QAEB=ΔU+WAEB ...(iii)
The change in internal energy in all the three cases is same i.e ΔU for each process.

WACB is +ve because area under curve during expansion (CB) is greater than area under curve during compression(AC)

WAEB is ve because area under curve during expansion (AE) is less than area under curve during compression(EB)

Hence on comparing (i), (ii), (iii) we get,
QACB>QADB>QAEB

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