Q.

Starting at temperature 300 K , one mole of an ideal , diatomic gas γ=1.4 is first compressed adiabatically from volume V1 to V2=V116 . It is then allowed to expand isobarically to volume 2V2. If all the processes are the quasi-static then the final temperature of the gas (in K ) is (to the nearest integer)___

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answer is 1819.

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Detailed Solution

During adiabatic processTVγ−1=constant    ⇒T1V1γ−1=T2.V2γ−1    ⇒T2=T1(V1V2)γ−1=300(16)75−1=T2=300(24)2/5  =T2=300×24×25During isobaric processV=nRTP     ⇒V∝T     ⇒V2V3=T2T3 ⇒V22V2=300×24×25T3     ⇒T3=2×300×28/5     =1818.85  =1819 K
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