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JEE Main 201910 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
10 April
Shift
Morning Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2019 · 10 April · Morning Shift · Physics · Thermodynamics

n moles of an ideal gas with constant volume heat capacity C _ v...

nn moles of an ideal gas with constant volume heat capacity Cv{C}_{v} undergo an isobaric expansion by certain volume. The ratio of the work done in the process, to the heat supplied is:

Options

  1. A4nRCv+nR\dfrac{4nR}{{C}_{v} + nR}
  2. B4nRCv−nR\dfrac{4nR}{{C}_{v} - nR}
  3. CnRCv+nR\dfrac{nR}{{C}_{v} + nR}
  4. DnRCv−nR\dfrac{nR}{{C}_{v} - nR}

Correct answer

C. nRCv+nR\dfrac{nR}{{C}_{v} + nR}

Step-by-step solution

From First Law of thermodynamics,
\DeltaQ=\DeltaU+\Deltaw\DeltaQ = \DeltaU + \Deltaw
\Deltaw\DeltaQ=1−\DeltaU\DeltaQ=1−ncv\DeltaTncp\DeltaT\dfrac{\Deltaw}{\DeltaQ} = 1 - \dfrac{\DeltaU}{\DeltaQ} = 1 - \dfrac{n{c}_{v}\DeltaT}{n{c}_{p}\DeltaT}
=1−cvR+cv= 1 - \dfrac{{c}_{v}}{R + {c}_{v}}
=RR+cv= \dfrac{R}{R + {c}_{v}}
As, ncv=Cvn{c}_{v} = {C}_{v}
\Deltaw\DeltaQ=nRCv+nR\dfrac{\Deltaw}{\DeltaQ} = \dfrac{nR}{{C}_{v} + nR}
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