Q 11-12-155JEE MainJEE Main 2017 (2 Apr)Medium
The temperature of an open room of volume $30\ \text{m}^3$ increases from $17^\circ$C to $27^\circ$C due to the sunshine. The atmospheric pressure in the room remains $1\times10^5$ Pa. If $n_i$ and $n_f$ are the number of molecules in the room before and after heating, then $n_f - n_i$ will be:
Answer: (A) $-2.5\times10^{25}$
The room is open, so $P$ and $V$ stay fixed while molecules leave. From $PV = nkT$, $n = \dfrac{PV}{kT}$:
$$n_f - n_i = \frac{PV}{k}\left(\frac{1}{300} - \frac{1}{290}\right) = \frac{(10^5)(30)}{1.38\times10^{-23}}\left(-\frac{10}{87000}\right)$$
$$= 2.17\times10^{29}\times(-1.15\times10^{-4}) \approx -2.5\times10^{25}$$
Solution by Sreeraj P, M.Sc Physics