Thermodynamics

JEE Main level, numerical answer, medium

137 of 190

Question 137 of 190

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 V1{V}_{1} to V2=V116.{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) ___________.

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Answer: 1819

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

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