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+1 vote
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in Chemistry by (53.0k points)

A reversible cyclic process for an ideal gas is shown below. Here, P , V and T are pressure , volume and temperature , respectively. The thermodynamic parameters q, w, H and U are heat, work, enthalpy and internal energy, respectively.

The correct option(s) is (are)
(A) qAC = ΔUBC and wAB = P2 (V2 – V1)
(B) wBC = P2 (V2 – V1) and qBC = ΔHAC
(C) ΔHCA < ΔUCA and qAC = ΔUBC
(D) qBC = ΔHAC and ΔHCA > ΔUCA

1 Answer

+2 votes
by (266k points)
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Best answer

Answer: (B,C)
Solution:

 AC → Isochoric
AB →Isothermal
BC →Isobaric
# qAC =  ΔUBC = nCV (T2 – T1)
WAB = nRT1 ln (V2/V1) A (wrong)
# qBC = ΔHAC = nCP (T2 – T1)
WBC = – P2(V1 – V2)   B (correct)
# nCP (T1–T2) < nCV(T1 – T2) C (correct)
ΔHCA < ΔUCA
# D (wrong)

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