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QJEEAdv19Ph1090: JEE Advanced 2019 Physics - Class 11 JEEAdv19Ph1090 Solution: One mole of a monatomic ideal gas goes through a thermodynamic…

One mole of a monatomic ideal gas goes through a thermodynamic cycle, as shown in the volume versus temperature (V-T) diagram. The correct statement(s) is/are [R is the gas constant]

JEE Advanced 2019 Paper 1 Offline Physics - Heat and Thermodynamics Question 41 English

Multiple Choice Options

  • A Work done in this thermodynamic cycle (1 →\to 2 →\to 3 →\to 4 →\to 1) is ∣W∣=12RT0\left| W \right| = {1 \over 2}R{T_0} .
  • B The ratio of heat transfer during processes 1 →\to 2 and 2 →\to 3 is ∣Q1→2Q2→3∣=53\left| {{{{Q_{1 \to 2}}} \over {{Q_{2 \to 3}}}}} \right| = {5 \over 3}
  • C The above thermodynamic cycle exhibits only isochoric and adiabatic processes.
  • D The ratio of heat transfer during processes 1 →\to 2 and 3 →\to 4 is ∣Q1→2Q3→4∣=12\left| {{{{Q_{1 \to 2}}} \over {{Q_{3 \to 4}}}}} \right| = {1 \over 2}

Step-by-Step Verified Solution

P-V graph of the given V-T graph is shown below.

JEE Advanced 2019 Paper 1 Offline Physics - Heat and Thermodynamics Question 41 English Explanation

(d) ∣ΔQ1→2ΔQ3→4∣=∣NCpΔT1→2NCpΔT3→4∣=T0T0/2=2\left| {{{\Delta {Q_{1 \to 2}}} \over {\Delta {Q_{3 \to 4}}}}} \right| = \left| {{{N{C_p}\Delta {T_{1 \to 2}}} \over {N{C_p}\Delta {T_{3 \to 4}}}}} \right| = {{{T_0}} \over {{T_0}/2}} = 2

(b) ∣ΔQ1→2ΔQ2→3∣=∣NCpΔT1→2NCpΔT3→4∣=CpCV=53\left| {{{\Delta {Q_{1 \to 2}}} \over {\Delta {Q_{2 \to 3}}}}} \right| = \left| {{{N{C_p}\Delta {T_{1 \to 2}}} \over {N{C_p}\Delta {T_{3 \to 4}}}}} \right| = {{{C_p}} \over {{C_V}}} = {5 \over 3}

(a) Wcycle=p0V0=nR[T02]{W_{cycle}} = {p_0}{V_0} = nR\left[ {{{{T_0}} \over 2}} \right]

Note For ideal gas equation,

pV = nRT

∴\therefore (pV)4 = (nRT)4

or p0V0 = nRT02nR{{{T_0}} \over 2} = RT02R{{{T_0}} \over 2}

as n = 1

(c) Wrong as no adiabatic process is involved.

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