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Question

# One mole of an ideal gas (monoatomic) is taken from A to C along the path ABC. The temperature of the gas is To at A. Then for the process ABC

A
Work done by gas is RTo
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B
Change in internal energy is 112RTo
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C
Heat absorbed by gas is 112RTo
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D
Heat absorbed by gas is 132RTo
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Solution

## The correct options are A Work done by gas is RTo C Heat absorbed by gas is 112RToWork done by gas for AB = Area under the line AB = Po(2Vo−Vo)=PoVo For 1 mole of an ideal gas, PoVo=RTo Work done by gas for process BC = 0 (∵ Volume is constant) Therefore, work done by gas for ABC = Work done by gas for AB + Work done by gas for BC Thus, total work done = RTo Internal energy change for ABC can be calculated by adding the internal energy change for AB and BC or directly between A and C because internal energy is point function which depends on final and initial state. Thus, internal energy change for AC, ΔU=nCvdt=nCv(TC−TA) For an ideal monoatomic gas, Cv=32R Now AB is a constant pressure process, thus V∝T ∴VBVA=TBTA Now VB=2Vo, VA=Vo and TA=To ⇒TB=2To Now BC is a constant volume process, thus P∝T ∴PCPB=TCTB Now PB=Po, PC=2Po and TB=2To ⇒TC=4To Thus, internal energy for ABC, ΔU=nCv(TC−TA) ΔU=1×32R(4To−To)=92RTo Since Q=U+W For ABC, Q=92RTo+RTo=112RTo

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