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JEE Advanced 2016Paper 2
Exam
JEE Advanced
Year
2016
Paper
Paper 2
Subject
Physics
Chapter
Thermodynamics

JEE Advanced 2016 · Paper 2 · Physics · Thermodynamics

A gas is enclosed in a cylinder with a movable frictionless piston....

A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure Pi=105PaandvolumeVi=10−3m3{P}_{i} = {10}^{5}Paandvolume{V}_{i} = {10}^{- 3}{m}^{3} changes to a final state at Pf=(132)×105PaandVf=8×10−3m3{P}_{f} = \left(\dfrac{1}{32}\right) \times {10}^{5}Paand{V}_{f} = 8 \times {10}^{- 3}{m}^{3} in an adiabatic quasi - static process, such that P3V5={P}^{3}{V}^{5} = constant. Consider another thermodynamic process that brings the system from the same initial state to the same final state in two steps: an isobaric expansion at Pi{P}_{i} followed by an isochoric (isovolumetric) process at volumes Vf{V}_{f} . The amount of heat supplied to the system in the two step process is approximatelyA gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure P _ i =

Options

  1. A112 J
  2. B294 J
  3. C588 J
  4. D813 J

Correct answer

C. 588 J

Step-by-step solution

In adiabatic process
P3V5={P}^{3}{V}^{5} = constant
⇒PV53=⇒ P{V}^{\dfrac{5}{3}} = constant
⇒γ=53⇒CV=32RandCP=52R⇒ \gamma = \dfrac{5}{3} ⇒ {C}_{V} = \dfrac{3}{2}Rand{C}_{P} = \dfrac{5}{2}R
In another process
A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure P _ i =
∆Q=nCP∆T+nCv∆T∆ Q = n{C}_{P} ∆ T + n{C}_{v} ∆ T
=52nR(TB−TA)+32nR(TC−TB)= \dfrac{5}{2}nR\left({T}_{B} - {T}_{A}\right) + \dfrac{3}{2}nR\left({T}_{C} - {T}_{B}\right)
∆Q=52(PBVB−PAVA)+32(PCVC−PBVB)∆ Q = \dfrac{5}{2}\left({P}_{B}{V}_{B} - {P}_{A}{V}_{A}\right) + \dfrac{3}{2}\left({P}_{C}{V}_{C} - {P}_{B}{V}_{B}\right)
Putting values
∆Q=587.5J≈588J∆ Q = 587.5J \approx 588J

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