Q.

One mole of monoatomic gas is taken through cyclic process shown below. TA = 300 K. Process AB is defined as PT = constant.

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a

Work done in process AB is - 400 R.

b

Change in internal energy in process CA is 900 R

c

Heat transferred in the process BC is 2000 R.

d

Change in internal energy in process CA is - 900 R.

answer is A.

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Detailed Solution

(a) Process AB :          PT= constant ⇒ nRT2V= constant    W=∫AB PdV=∫AB Constant TdV⇒ dVdT=2nRT Constant       W=∫300100 Constant T⋅2nRT Constant dt⇒ PAPB=TBTA⇒ 13=TBTA⇒ TB=3003=100⇒ W=2nR(100−300)⇒ WAB=−400nR(b) Process CA : Isochoric, so PT is constant :             TATC=PAPC        TATC=PAPB⇒ TATC=13⇒ TC=3TA           TC=900R⇒ ΔU=nCVΔT                =(1)32R×TA−TC                =32R×(300−900)                =32R×−600=−900R⇒ |ΔU|=(900R)(c) Process BC:  Isobaric          Q=nCPΔT⇒Q=(1)52R×TC−TB    Q=52R×(900−100)⇒Q=52R×800⇒Q=2000R
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