Question bank/JEE Main/Physics

JEE MainPhysicsToughThermodynamics

A system undergoes three quasi-static processes sequentially as
indicated in the given figure.
1→21 \rightarrow 2is an isobaric process,
2→32 \rightarrow 3 is a polytrophic process with
γ=43\gamma = \frac{4}{3}and 3→13 \rightarrow 1 is a process in which PV=
constant.

P2=P1=4×105Nm−2,P3=1×105Nm−2P_{2} = P_{1} = 4 \times 10^{5}Nm^{- 2},P_{3} = 1 \times 10^{5}Nm^{- 2}and V1=1m3\ V_{1} = 1m^{3}.
The heat transfer of the cycle is
ΔQ,\Delta Q, change in internal
energy is
ΔU\Delta U and the work done is ΔW\Delta W. for polytropic
process
PVγ=PV^{\gamma} = constant.
(Take
l\mathcal{l}n2 = 0.693). Then

Show answer and solution
For the process 3→13 \rightarrow 1
P1V1=P3V3⇒V3=P1V1P3=4×105×11×105=4m3P_{1}V_{1} = P_{3}V_{3} \Rightarrow V_{3} = \frac{P_{1}V_{1}}{P_{3}} = \frac{4 \times 10^{5} \times 1}{1 \times 10^{5}} = 4m^{3}
For the process
2→32 \rightarrow 3
P2V2γ=P3V3γ⇒V2=(P3P2)1γV3=4(14)34=414=2m3P_{2}V_{2}^{\gamma} = P_{3}V_{3}^{\gamma} \Rightarrow V_{2} = \left( \frac{P_{3}}{P_{2}} \right)^{\frac{1}{\gamma}}V_{3} = 4\left( \frac{1}{4} \right)^{\frac{3}{4}} = 4^{\frac{1}{4}} = \sqrt{2}m^{3}
For the process
1→21 \rightarrow 2
W12=P1(V2−V1)=(2−1)4×105JW_{12} = P_{1}\left( V_{2} - V_{1} \right) = \left( \sqrt{2} - 1 \right)4 \times 10^{5}J
For the process
2→32 \rightarrow 3
W23=(P2V2−P3V3)(γ−1)=12(2−1)×105JW_{23} = \frac{\left( P_{2}V_{2} - P_{3}V_{3} \right)}{(\gamma - 1)} = 12(\sqrt{2} - 1) \times 10^{5}J
For the process
3→13 \rightarrow 1,
w31=−P1V1lnV3V1−4×105×1×ln(4)w_{31} = - P_{1}V_{1}ln\frac{V_{3}}{V_{1}} - 4 \times 10^{5} \times 1 \times ln(4)
For cyclic process,
Δ U=0\Delta\ U = 0
∴ΔQ=ΔU+ΔW=ΔW=4×105(2−1)\therefore\Delta Q = \Delta U + \Delta W = \Delta W = 4 \times 10^{5}(\sqrt{2} - 1)
1→2→3→11 \rightarrow 2 \rightarrow 3 \rightarrow 1
ΔQ=+12×105(2−1)−4×105×2×0.693\Delta Q = + 12 \times 10^{5}(\sqrt{2} - 1) - 4 \times 10^{5} \times 2 \times 0.693
=1.08×105J= 1.08 \times 10^{5}J

Source: Grass-G / P-08 Physics paper 23 October.docx

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