$32$ gram of oxygen gas at temperature $27^{o} C$ is compressed adiabatically to $1 / 3$ of its change in internal energy. $γ = 1.5$ of oxygen)

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Solution

Given $T_{1} = 300 K$ $\frac{V_{2}}{V_{1}} = \frac{1}{3}$ $γ = 1.5$
$\Rightarrow C_{V}$ for $O_{2} = \frac{5}{2} R$
$\Rightarrow$ Change in internal energy $= d U = C_{V} d T$
$= C_{V} (T_{2} - T_{1}) \to (1)$
For adiabatic compression,
$\Rightarrow T V^{γ - 1} =$ Constant So, $T_{2} - T_{1} = (\sqrt{3} - 1) T_{1}$
$\Rightarrow T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1}$ $= (\sqrt{3} - 1) 300 K$
$\Rightarrow \frac{T_{2}}{T_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ - 1} = {(3)}^{1.5 - 1} = \sqrt{3}$ $d U = \frac{5}{2} R (\sqrt{3} - 1) 300 K$
$= d U = 4564.7 J .$
Hence, solve.

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