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,d s fp=kuqlkj rhu vnZ~/k&fLFkj çØeks ls xqtkjk A
vkarfjd












Show answer and solution
For the process izØe ds fy, 3 → 1
[IMAGE] For the processizØe ds fy, 2→ 3

[IMAGE] For the processizØe ds fy, 1 → 2

[IMAGE] For the processizØe ds fy, 2 → 3

[IMAGE] For the process 3 → 1

[IMAGE] For cyclic process pØh

=1.08×105J= 1.08 \times 10^{5}J






Source: Grass-G / Ph-25 Physics paper 7 Jan.docx

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