JEE Main & Advanced Physics Thermodynamical Processes Sample Paper Topic Test - Thermodynamical Processes

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

Correct Answer: A

Solution :

[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 \[\Delta U=\Delta Q-W=0-W=46kJ\]