Thermodynamic processes

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Question

An engine takes in 5 moles of air at 20ºC and 1atm, and compresses it adiabatically to 1/10th of the original volume. Assuming air to be a diatomic ideal gas made up of rigid molecules, the change in its internal energy during this process comes out to be X kJ. The value of X to the nearest integer is________

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Solution

$T_{i n} = 10 ° C = 10 + 273 = 293 K$
$P V^{γ} = c o n s \tan t, γ = \frac{7}{5}$
$\Rightarrow T V^{γ - 1} = c o n t \Rightarrow T V^{\frac{7}{5} - 1} = c o n t$
$\Rightarrow T V^{\frac{7 - 5}{5}} = c o n t \Rightarrow T V^{\frac{2}{5}} = c o n t$
$\Rightarrow T_{f i n} {(\frac{V}{10})}^{2 / 5} = T_{i n} V^{2 / 5}$
$\Rightarrow T_{f i n} = 293 (10^{2 / 5}) (∵ T_{i n} = 10 ° C = 10 + 273 = 293 K)$
$Δ U = \frac{n R Δ T}{γ - 1} = \frac{5 R}{2 / 5} \times 293 (10^{2 / 5} - 1) = \frac{25 R}{2} (10^{2 / 5} - 1) \times 293 = \frac{25 R}{2} (2.511 - 1) \times 293 = 12.5 R (1.511) \times 293 = 46 k J$

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