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QJEEAdv19Phcd65: JEE Advanced 2019 Physics - Class 11 JEEAdv19Phcd65 Solution: If the process carried out on one mole of monoatomic ideal gas…

If the process carried out on one mole of monoatomic ideal gas is as shown in the PV-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 40 English

Multiple Choice Options

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

Step-by-Step Verified Solution

(I) W1→2→3=W1→2+W2→3=P0[2V0−V0]+0=P0V0{W_{1 \to 2 \to 3}} = {W_{1 \to 2}} + {W_{2 \to 3}} = {P_0}[2{V_0} - {V_0}] + 0 = {P_0}{V_0}

W1→2→3=P0V0=RT03{W_{1 \to 2 \to 3}} = {P_0}{V_0} = {{R{T_0}} \over 3} (I →\to Q)

U1→2→3=32[3P02×2V0−P0V0]=32×2P0V0=3P0V0=RT0{U_{1 \to 2 \to 3}} = {3 \over 2}\left[ {{{3{P_0}} \over 2} \times 2{V_0} - {P_0}{V_0}} \right] = {3 \over 2} \times 2{P_0}{V_0} = 3{P_0}{V_0} = R{T_0} (II →\to R)

(III) Q1→2→3=U1→2→3+W1→2→3=RT0+RT03=4RT03{Q_{1 \to 2 \to 3}} = {U_{1 \to 2 \to 3}} + {W_{1 \to 2 \to 3}} = R{T_0} + {{R{T_0}} \over 3} = {{4R{T_0}} \over 3} (III →\to S)

(IV) Q1→2=nCpΔT=n52R(T2−T1)=52[P02V0−P0V0]=52P0V0=52RT03=56R0T0{Q_{1 \to 2}} = n{C_p}\Delta T = n{5 \over 2}R({T_2} - {T_1}) = {5 \over 2}[{P_0}2{V_0} - {P_0}{V_0}] = {5 \over 2}{P_0}{V_0} = {5 \over 2}{{R{T_0}} \over 3} = {5 \over 6}{R_0}{T_0} (IV →\to U)

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