Question

The temperature of an open room of volume $30m^3$ increases from ${17}^{\circ }C$ to ${27}^{\circ }C$ due to the sunshine. The atmospheric pressure in the room remains $1\times {10}^{5}Pa$. 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

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

Answer 1

The correct option is D $\begin{array}{c}-2.5\times {10}^{25}\end{array}$

Using ideal gas equation,
$PV=NRT$
($N$ is number of moles)
$P_0{V}_0=N_{i}R\times 290$ ------------- (1)
After heating,
$P_0{V}_0=N_{f}R\times 300$ --------- (2)
From equation (1) and (2)
$N_f-N_i=\frac{P_o{V}_0}{R\times 300}-\frac{P_o{V}_0}{R\times 290}$
= $\frac{-P_0{V}_0}{R}[\frac{10}{290\times 300}]$
Hence $n_f-n_i=\frac{-P_0{V}_0}{R}\times [\frac{10}{290\times 300}]$$\times 6.023\times {10}^{23}$
Putting $P_0={10}^5{P}_A$ and $V_0=30m^3$
Number of molecules $n_f-n_i=-2.5\times {10}^{25}$