An engine takes in 5 moles of air at 20oC and 1 atm, and compresses it adiabaticaly to 1/10th of the orginal 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 __________
Option: 1 11
Option: 2 46
Option: 3 35
Option: 4 28

Answers (1)

 \begin{aligned} &\text {Given Gas is Diatomic }\\ &So \ \mathrm{f}=5\\ &\gamma=7 / 5\\ &\mathrm{T}_{\mathrm{i}}=\mathrm{T}=273+20=293 \mathrm{~K}\\ &\mathrm{V}_{\mathrm{i}}=\mathrm{V}\\ &\mathrm{V}_{\mathrm{f}}=\mathrm{V} / 10 \end{aligned}

For adiabatic process

\begin{aligned} &\mathrm{TV}^{\gamma-1}=\text { constant }\\ &\mathrm{T}_{1} \mathrm{~V}_{1}^{\gamma-1}=\mathrm{T}_{2} \mathrm{~V}_{2}^{\gamma-1}\\ &\mathrm{T.V}^{7 / 5-1}=\mathrm{T}_{2}\left(\frac{\mathrm{V}}{10}\right)^{7 / 5-1} \end{aligned}

\Rightarrow \mathrm{T}_{2}=\mathrm{T} *10^{2 / 5}

So

 \begin{array}{l} \Rightarrow T_{2}=293(10)^{2 / 5}=735.98=736 \mathrm{~K} \\ \therefore \Delta T=443 \mathrm{~K} \end{array}

Now

\begin{aligned} \Delta U &=\frac{\operatorname{nfR}\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)}{2}=\frac{5 \times 5 \times \frac{25}{3} \times 443}{2} =46.038 * 10 ^3 \ J \approx 46 \mathrm{~kJ} \end{aligned}

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