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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