Question

The temperature of an open room of volume $30{\text{m}}^3$ increases from $17{}^{\circ }\text{C}$ to $27{}^{\circ }\text{C}$ due to sunshine. The atmospheric pressure in the room remains $1\times {10}^5\text{Pa}$. If $n_i$ and $n_f$ are the numbers of molecules in the room before and after heating, then $n_f-n_i$ will be

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

Answer 1

The correct option is B $-2.5\times {10}^{25}$

Given:-
$T_i=17{}^{\circ }\text{C}=290{}^{\circ }\text{K}$
$T_f=27{}^{\circ }\text{C}=300{}^{\circ }\text{K}$
$V_0=30{\text{m}}^3;P_0={10}^5\text{Pa}$

Using ideal gas equation,

$P_0{V}_0=N_{i}R{T}_i=N_{f}R{T}_f$

Difference between number of moles,

$N_f-N_i=\frac{P_0{V}_0}{R\times 300}-\frac{P_0{V}_0}{R\times 290}$

As we know that, $n=\frac{N}{N_A}=\frac{N}{6.023\times {10}^{23}}$

$\Rightarrow n_f-n_i=\frac{-30\times {10}^5\times 10\times 6.023\times {10}^{23}}{8.314\times 290\times 300}$

$\therefore n_f-n_i=-2.5\times {10}^{25}$

Hence, option $\left(\text{B}\right)$ is correct.