Skip to content
dronacharya.ai
← JEE Advanced 2014 · Paper 1
PhysicsInteger+3Question 20 of 60

JEE Advanced 2014 · Paper 1, Physics question 20

A thermodynamic system is taken from an initial state ii with internal energy Ui=100U_i=100 J to the final state ff along two different paths iafiaf and ibfibf, as schematically shown in the figure. The work done by the system along the paths afaf, ibib and bfbf are Waf=200W_{af}=200 J, Wib=50W_{ib}=50 J and Wbf=100W_{bf}=100 J respectively. The heat supplied to the system along the path iafiaf, ibib and bfbf are QiafQ_{iaf}, QibQ_{ib} and QbfQ_{bf} respectively. If the internal energy of the system in the state bb is Ub=200U_b=200 J and Qiaf=500Q_{iaf}=500 J, the ratio Qbf/QibQ_{bf}/Q_{ib} is

P-V diagram: path i→a is vertical (isochoric, pressure rising), a→f horizontal (isobaric); the other path goes i→b horizontally then b→f along a rising curve to f.
Answer and worked solution

Official answer: 2

Path iaf. iaia is isochoric (W=0W=0), so Wiaf=200W_{iaf}=200 J:
Qiaf=Uf−Ui+Wiaf  ⟹  500=Uf−100+200  ⟹  Uf=400 J.Q_{iaf}=U_f-U_i+W_{iaf}\implies500=U_f-100+200\implies U_f=400\text{ J}.

Path ib. Qib=(200−100)+50=150Q_{ib}=(200-100)+50=150 J.

Path bf. Qbf=(400−200)+100=300Q_{bf}=(400-200)+100=300 J.

Ratio. Qbf/Qib=300/150=2Q_{bf}/Q_{ib}=300/150=2.

Answer: 2