Answer: Option C
:
C
The change in internal energy $\Delta E_{int}=n{C}_V\Delta T$. We need to obtain $C_V$ from the given information. Now,
$\gamma =\left[\frac{C_P}{C_V}\right]$, and, $C_P-C_V=R$
$\Rightarrow \gamma =\left[\frac{C_V+R}{C_V}\right]=[1+\frac{R}{C_V}]$
$\Rightarrow \frac{R}{C_V}=\gamma -1$
$\Rightarrow C_V=\left[\frac{R}{\gamma -1}\right]$
Therefore,
$\Delta E_{int}=n{C}_V\Delta T$
$=2500\times \left(\frac{8.314}{1.4-1}\right)\times (26-35)J$
$=-4.68\times {10}^{5}J$.
It's a negative change, meaning the internal energy has decreased with cooling, which is consistent with theory. In fact, if an AC had to do this cooling, its job would be take $4.68\times {10}^{5}J$from the internal energy in the air and dump it outside. This will be dealt with in greater detail when we study the second law of thermodynamics.