Question

The temperature of an open room of volume $30{\mathrm{m}}^3$ increases from ${17}^{\mathrm{o}}\mathrm{C}\mathrm{to}{27}^{\mathrm{o}}\mathrm{C}$ due to the sunshine. The atmospheric pressure in the room remains $1\times {10}^5\mathrm{Pa}$. If ${\mathrm{n}}_{\mathrm{i}}\mathrm{and}{\mathrm{n}}_{\mathrm{f}}$ are the number of molecules in the room before and after heating, then ${\mathrm{n}}_{\mathrm{f}}-{\mathrm{n}}_{\mathrm{i}}$ will be:

• A. $-1.61\times {10}^{23}$
• B. $1.38\times {10}^{23}$
• C. $2.5\times {10}^{25}$
• D. $-2.5\times {10}^{25}$

Answer 1

The correct option is D

$-2.5\times {10}^{25}$

The explanation for the correct option:

D: $-2.5\times {10}^{25}$

The calculation to determine the change in the number of molecules:

$\begin{array}{rcl}{\mathrm{n}}_1& =& \mathrm{initial}\mathrm{number}\mathrm{of}\mathrm{moles}\\ {\mathrm{n}}_1& =& {\mathrm{P}}_1{\mathrm{V}}_1/{\mathrm{RT}}_1=({10}^5\times 30)/(8.3\times 290)\\ & =& 1.24\times {10}^3\\ {\mathrm{n}}_2& =& \mathrm{final}\mathrm{number}\mathrm{of}\mathrm{moles}{\mathrm{n}}_2={\mathrm{P}}_2{\mathrm{V}}_2/{\mathrm{RT}}_2\\ & =& ({10}^5\times 30)/(8.3\times 300)\\ & =& 1.20\times {10}^3\\ & & \mathrm{Change}\mathrm{of}\mathrm{number}\mathrm{of}\mathrm{molecules}:\\ \mathrm{nf}–\mathrm{ni}& =& (\mathrm{n}2–\mathrm{n}1)\times 6.023\times {10}^{23}\\ & =& -2.5\times 1025\end{array}$

Therefore, option D $-2.5\times {10}^{25}$ is the correct option.