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JEE Advanced 2019Paper 1
Exam
JEE Advanced
Year
2019
Paper
Paper 1
Subject
Physics
Chapter
Thermodynamics

JEE Advanced 2019 · Paper 1 · Physics · Thermodynamics

One mole of a monoatomic ideal gas goes through a thermodynamic...

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

Options

  1. AThe ratio of heat transfer during processes 1→2and1 \to 2 and2 \to 3is∣Q1→2Q2→3∣=53is \left|\dfrac{{Q}_{1 \to 2}}{{Q}_{2 \to 3}}\right| = \dfrac{5}{3}
  2. BThe ratio of heat transfer during processes 1→2and1 \to 2 and3 \to 4is∣Q1→2Q3→4∣=12is \left|\dfrac{{Q}_{1 \to 2}}{{Q}_{3 \to 4}}\right| = \dfrac{1}{2}
  3. CThe above thermodynamic cycle exhibits only isochoric and adiabatic processes.
  4. DWork done in this thermodynamic cycle (1→2→3→4→1)\left(1 \to 2 \to 3 \to 4 \to 1\right) is \left|W\right| = \dfrac{1}{2}R{T}_{0}

Correct answer

A. The ratio of heat transfer during processes 1→2and1 \to 2 and2 \to 3is∣Q1→2Q2→3∣=53is \left|\dfrac{{Q}_{1 \to 2}}{{Q}_{2 \to 3}}\right| = \dfrac{5}{3}

Step-by-step solution

One mole of a monoatomic ideal gas goes through a thermodynamic cycle, as shown in the volume versus temperature (V -
(A)⇒∣ΔQ1→2ΔQ2→3∣=∣NCPΔT1→2NCVΔT2→3∣=CPCV=53⇒\left(A\right) ⇒ \left|\dfrac{\Delta{Q}_{1 \to 2}}{\Delta{Q}_{2 \to 3}}\right| = \left|\dfrac{N{C}_{P}\Delta{T}_{1 \to 2}}{N{C}_{V}\Delta{T}_{2 \to 3}}\right| = \dfrac{{C}_{P}}{{C}_{V}} = \dfrac{5}{3} ⇒ correct
\left(B\right) ⇒ \left|\dfrac{\Delta{Q}_{1 \to 2}}{\Delta{Q}_{3 \to 4}}\right| = \dfrac{{T}_{0}}{{T}_{0} / 2} = 2 ⇒Incorrect<br>Incorrect<br >( C ) ⇒NoadiabaticprocessisinvolvedNo adiabatic process is involved⇒incorrect<br>incorrect<br >( D ) ⇒Workdone=areainsideWork done = area insideP - Vdiagramdiagram= {P}_{0}{V}_{0} = nR\dfrac{{T}_{0}}{2}[atpoint[at point4indiagram]in diagram]⇒correct.<br>correct.<br >1 \to 2 ⇒isobaricatpressureisobaric at pressure2{P}_{0}<br><br >2 \to 3 ⇒isobaricatvolumeisobaric at volume2{V}_{0}<br><br >3 \to 4 ⇒isobaricatpressureisobaric at pressure{P}_{0}<br><br >4 \to 1 ⇒isobaricatvolumeisobaric at volume{V}_{0}<br>Given,numberofmoles,<br >Given, number of moles,n = 1<br>and<br >andr = \dfrac{{C}_{P}}{{C}_{V}} = \dfrac{5}{3}(formonoatomicgas)<br>Now,(for mono atomic gas)<br >Now,\Delta{Q}_{1 \to 2}{=}^{n}{C}_{P} ( {T}_{2} - {T}_{1} )<br><br >= n ⋅ {C}_{p} ⋅ {T}_{0}<br><br >\Delta{Q}_{2 \to 3}{=}^{n}{C}_{V}\left({T}_{3} - {T}_{2}\right)=n⋅CV(−T2)= n ⋅ {C}_{V}\left(- {T}_{2}\right)
ΔQ3→4=n⋅CP(T4−T3)=n⋅CP(−T02)\Delta{Q}_{3 \to 4} = n ⋅ {C}_{P}\left({T}_{4} - {T}_{3}\right) = n ⋅ {C}_{P}\left(\dfrac{- {T}_{0}}{2}\right)
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