• # question_answer A diesel engine takes in 5 moles of air at $20{}^\circ C$ and 1 atm, and compresses it adiabatically to $\frac{1}{10}\text{th}$ of the original volume. If air is diatomic then work done and change in internal energy is A) - 46 kJ, 46 kJ B) 36 kJ, - 36 kJ C) 46 kJ, - 46 kJ D) - 36 kJ, 36 kJ

 [a]Let ${{p}_{1}}=1$ atm, $n=5\,mol,\text{ }293K$ ${{V}_{2}}=\frac{{{V}_{1}}}{10}$ Using   ${{T}_{1}}V_{1}^{\gamma -1}={{T}_{2}}V_{2}^{\gamma -1}$ $\Rightarrow$            ${{T}_{2}}={{T}_{1}}{{\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{\gamma -1}}$ $=293{{(10)}^{0.4}}=736K$ Now, work done $=\frac{nR({{T}_{1}}-{{T}_{2}})}{\gamma -1}$ $=\frac{5\times 8.3\times (293-736)}{0.4}=-46kJ$ and $\Delta U=\Delta Q-W=0-W=46kJ$