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