The temperature of an open room of volume 30 m$$^{3}$$ increases from 17°C to 27°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:

We start with the ideal­-gas equation written for a collection of molecules:

$$PV = NkT$$

Here $$P$$ is the pressure, $$V$$ is the volume, $$N$$ is the number of molecules, $$k = 1.38 \times 10^{-23}\,\text{J K}^{-1}$$ is Boltzmann’s constant and $$T$$ is the absolute temperature in kelvin.

The room is open to the atmosphere, so the pressure inside always equals the atmospheric pressure and remains constant at

$$P = 1 \times 10^{5}\,\text{Pa}.$$

The rigid walls keep the volume fixed at

$$V = 30\ \text{m}^{3}.$$

Because both $$P$$ and $$V$$ are constant, the product $$PV$$ is constant, and we can write for two different states:

$$PV = N_i k T_i \quad\text{and}\quad PV = N_f k T_f.$$

Dividing the two equations gives

$$\frac{N_f}{N_i} = \frac{T_i}{T_f}.$$ So

$$N_f = N_i\,\frac{T_i}{T_f}.$$

We want the difference $$N_f - N_i$$, so we first express both $$N_i$$ and $$N_f$$ individually using $$PV = NkT$$.

For the initial state (17 °C):

Temperature in kelvin: $$T_i = 17^{\circ}\text{C} + 273 = 290\ \text{K}.$$

Hence

$$N_i = \frac{PV}{kT_i}.$$

For the final state (27 °C):

Temperature in kelvin: $$T_f = 27^{\circ}\text{C} + 273 = 300\ \text{K}.$$

Thus

$$N_f = \frac{PV}{kT_f}.$$

Now we write the required difference:

$$N_f - N_i = \frac{PV}{k}\left(\frac{1}{T_f} - \frac{1}{T_i}\right).$$

We substitute all the numerical values step by step.

First calculate the factor $$\dfrac{PV}{k}$$:

$$\frac{PV}{k} = \frac{(1 \times 10^{5}\,\text{Pa})(30\,\text{m}^{3})}{1.38 \times 10^{-23}\,\text{J K}^{-1}} = \frac{3.0 \times 10^{6}\,\text{J}}{1.38 \times 10^{-23}\,\text{J K}^{-1}} = 2.1739 \times 10^{29}\,\text{K}^{-1}.$$

Next evaluate the bracket containing the temperatures:

$$\frac{1}{T_f} - \frac{1}{T_i} = \frac{1}{300\ \text{K}} - \frac{1}{290\ \text{K}} = \frac{290 - 300}{300 \times 290} = \frac{-10}{87000} = -1.1494 \times 10^{-4}\,\text{K}^{-1}.$$

Multiplying the two results gives

$$N_f - N_i = \left(2.1739 \times 10^{29}\,\text{K}^{-1}\right) \left(-1.1494 \times 10^{-4}\,\text{K}^{-1}\right) = -2.50 \times 10^{25}.$$

The negative sign tells us that some molecules have left the room so that the pressure could stay constant while the temperature increased.

Hence, the correct answer is Option A.