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JEE Advanced · Physics · 13. Thermodynamics

One mole of a monoatomic ideal gas goes through a thermodynamic cycle, as shown in the volume versus temperature V-T diagram. The correct statement(s) is/are:
[ R is the gas constant]

  1. A The ratio of heat transfer during processes 12 and 23 is Q12Q23=53
  2. B The ratio of heat transfer during processes 12 and 34 is Q12Q34=12
  3. C The above thermodynamic cycle exhibits only isochoric and adiabatic processes.
  4. D Work done in this thermodynamic cycle 12341 is W=12RT0
Verified Solution

Answer & Solution

Correct Answer

(D) Work done in this thermodynamic cycle 12341 is W=12RT0

Step-by-step Solution

Detailed explanation


AΔQ12ΔQ23=NCPΔT12NCVΔT23=CPCV=53 correct
BΔQ12ΔQ34=T0T0/2=2 Incorrect
(C) No adiabatic process is involved incorrect
(D) Work done = area inside P-V diagram =P0V0=nRT02 [at point 4 in diagram] correct.
12 isobaric at pressure 2P0
23 isobaric at volume 2V0
34 isobaric at pressure P0
41 isobaric at volume V0
Given, number of moles, n=1
and r=CPCV=53 (for mono atomic gas)
Now, ΔQ12=nCP(T2-T1)
=nCpT0
ΔQ23=nCVT3-T2=nCV-T2
ΔQ34=nCPT4-T3=nCP-T02
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