QJEEM20Ph596c: JEE Main 2020 Physics - Class 11 JEEM20Ph596c Solution: Starting at temperature 300 K, one mole of an ideal diatomic…

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QJEEM20Ph596c: JEE Main 2020 Physics - Class 11 JEEM20Ph596c Solution: Starting at temperature 300 K, one mole of an ideal diatomic…

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

γ

= 1.4) is first compressed
adiabatically from volume V₁ to V₂ =

\frac{V_{1}}{16}

. 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 ^oK) is (to the nearest integer) _____.

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Accepted Answer

T₁V₁^{$γ$ –1} = T₂V₂ ^{$γ$ –1}

300 \times V^{\frac{7}{5} - 1} = T_{2} {(\frac{V}{16})}^{\frac{7}{5} - 1}

\Rightarrow

T₂ = 300 × (16)^0.4

Isobaric process

V =

\frac{n R T}{P}

V₂ = kT₂... (1)

2V ₂ = KT_f... (2)

T_f = 2T₂ = 300 × 2 × (16)^0.4 = 1818 K

QJEEM20Ph596c: JEE Main 2020 Physics - Class 11 JEEM20Ph596c Solution: Starting at temperature 300 K, one mole of an ideal diatomic…

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