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QJEEAdv19Ph707d: JEE Advanced 2019 Physics - Class 11 JEEAdv19Ph707d Solution: If the process on one mole of monatomic ideal gas is as shown…

If the process on one mole of monatomic ideal gas is as shown in the TV-diagram with P0V0=13RT0{P_0}{V_0} = {1 \over 3}R{T_0} , the correct match is,

JEE Advanced 2019 Paper 2 Offline Physics - Heat and Thermodynamics Question 39 English

Multiple Choice Options

  • A I →\to P, II →\to R, III →\to T, IV →\to S
  • B I →\to P, II →\to T, III →\to Q, IV →\to T
  • C I →\to S, II →\to T, III →\to Q, IV →\to U
  • D I →\to P, II →\to R, III →\to T, IV →\to P

Step-by-Step Verified Solution

1-2 process is isothermal and 2-3 process is isochoric.

(I) W1→2=nRI ln⁡VfVi=1×RT03 ln⁡V2V1{W_{1 \to 2}} = nRI\,\ln {{{V_f}} \over {{V_i}}} = 1 \times R{{{T_0}} \over 3}\,\ln {{{V_2}} \over {{V_1}}}

W2 →\to 3 = 0 (Isochoric process)

W1→2→3{W_{1 \to 2 \to 3}} = W1→2{W_{1 \to 2}} + W2→3{W_{2 \to 3}}

= RT03{{R{T_0}} \over 3} ln 2(I →\to P)

(II) ΔU=f2nR(Tf−Ti)\Delta U = {f \over 2}nR({T_f} - {T_i})

ΔU1→2→3=32[T0−T03]\Delta {U_{1 \to 2 \to 3}} = {3 \over 2}\left[ {{T_0} - {{{T_0}} \over 3}} \right]

=RT0= R{T_0} (II →\to R)

(III) Q1→2→3=ΔU1→2→3+W1→2→3{Q_{1 \to 2 \to 3}} = \Delta {U_{1 \to 2 \to 3}} + {W_{1 \to 2 \to 3}}

(First law of thermodynamics)

=RT0+RT03ln⁡2= R{T_0} + {{R{T_0}} \over 3}\ln 2

RT03[3+ln⁡2]{{R{T_0}} \over 3}[3 + \ln 2] (III →\to T)

(IV) Q1→2=ΔU1→2+W1→2=0+RT03ln2=RT03ln2{Q_{1 \to 2}} = \Delta {U_{1 \to 2}} + {W_{1 \to 2}} = 0 + {{R{T_0}} \over 3}ln2 = {{R{T_0}} \over 3}ln2 (IV →\to P)

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