QJEEM18Phaea7: JEE Main 2018 Physics - Class 11 JEEM18Phaea7 Solution: Two moles of an ideal monatomic gas occupies a volume V at 27 o …

For adiabatic process,

pv ^$\gamma$ = constant.

and we know, pv = nRT

$\therefore\,\,\,$ p = ${{nRT} \over v}$

$\therefore\,\,\,$ ${{nRT} \over v} \times {v^\gamma }$ = constant

$\Rightarrow$ $\,\,\,$ T v ^{$\gamma$ $-$ 1} = constant.

$\therefore\,\,\,$ T₁ v₁ ^{$\gamma$ $-$ 1} = T₂ v₂ ^{$\gamma$ $-$ 1}

Given that,

This is a monoatomic gas.

So, degree of frequency f = 3

$\therefore\,\,\,$ $\gamma$ = ${{{c_p}} \over {{c_v}}}$ = 1 + ${2 \over f}$ = 1 + ${2 \over 3}$ = ${5 \over 3}$

T₁ = 27 + 273 = 300 K

v₁ = v

and v₂ = 2v

$\therefore\,\,\,$ T₂ (2V) $^{{2 \over 3}}$ = 300(v) $^{{2 \over 3}}$

$\Rightarrow$ $\,\,\,$ T₂ = ${{300} \over {{2^{{2 \over 3}}}}}$ = 189 K

Change in internal energy,

$\Delta$ U = ${1 \over 2}$ nfR $\Delta$ T

= ${1 \over 2}$ $\times$ 2 $\times$ 3 $\times$ 8.31 $\times$ (189 $-$ 300)

= $-$ 2.7 KJ