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JEE Main 20209 JanuaryEvening Shift
Exam
JEE Main
Year
2020
Date
9 January
Shift
Evening Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2020 · 9 January · Evening Shift · Physics · Thermodynamics

Starting at temperature 300K, one mole of an ideal diatomic gas (γ...

Starting at temperature 300K,300K, one mole of an ideal diatomic gas (γ=1.4)\left(\gamma = 1.4\right) is first compressed adiabatically from volume V1toV2=V116.{V}_{1} to {V}_{2} = \dfrac{{V}_{1}}{16} . It is then allowed to expand isobarically to volume 2V22{V}_{2} . If all the processes are the quasi-static then the final temperature of the gas (in oK{}^{o}K ) is (to the nearest integer) ___________.Starting at temperature 300K, one mole of an ideal diatomic gas (γ = 1.4) is first compressed adiabatically from volume

Correct answer

0

Step-by-step solution

PVy=constantP{V}^{y} = constant
TV−1=constantT{V}^{- 1} = constant
300(V1)1.4−1=TB(V116)2/5300{\left({V}_{1}\right)}^{1.4 - 1} = {T}_{B}{\left(\dfrac{{V}_{1}}{16}\right)}^{2 / 5}
TB=300×28/5{T}_{B} = 300 \times {2}^{8 / 5}
Now for BC process
VBTB=VcTc\dfrac{{V}_{B}}{{T}_{B}} = \dfrac{{V}_{c}}{{T}_{c}}
Tc=VcTBvB=2×300×28/5{T}_{c} = \dfrac{{V}_{c}{T}_{B}}{{v}_{B}} = 2 \times 300 \times {2}^{8 / 5}
Tc=1818.859{T}_{c} = 1818.859
Tc=1819K{T}_{c} = 1819K
Starting at temperature 300K, one mole of an ideal diatomic gas (γ = 1.4) is first compressed adiabatically from volume
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