Answer :
The mass of air leaving the room due to temperature change is approximately 1311.783 grams.
To find the mass of air leaving the room, we can use the ideal gas law, which relates the pressure, volume, temperature, and number of moles of a gas:
[tex]\[ PV = nRT \][/tex]
Where:
- [tex]\( P \)[/tex] is the pressure of the gas (in Pascals),
- [tex]\( V \)[/tex] is the volume of the gas (in cubic meters),
- [tex]\( n \)[/tex] is the number of moles of the gas,
- [tex]\( R \) is the ideal gas constant (\( 8.314 \, \text{J/(mol} \cdot \text{K)} \)),[/tex]
- [tex]\( T \)[/tex] is the temperature of the gas (in Kelvin).
We can rearrange the ideal gas law to solve for the number of moles (\( n \)):
[tex]\[ n = \frac{PV}{RT} \][/tex]
Given:
[tex]- \( P = 126.6 \, \text{kPa} = 126600 \, \text{Pa} \) (converted to Pascals),[/tex]
[tex]- \( V = 123 \, \text{m}^3 \),[/tex]
[tex]- \( T_1 = 22.2°C = 22.2 + 273.15 \, \text{K} \),[/tex]
[tex]- \( T_2 = 31°C = 31 + 273.15 \, \text{K} \).[/tex]
First, let's calculate the number of moles of air in the room before and after the temperature change:
[tex]\[ n_1 = \frac{P \times V}{R \times T_1} \][/tex]
[tex]\[ n_2 = \frac{P \times V}{R \times T_2} \][/tex]
Then, we can find the difference in the number of moles, which represents the mass of air leaving the room:
[tex]\[ \Delta n = n_1 - n_2 \][/tex]
Finally, we can convert the change in moles to mass using the average molar mass of air (38.1 g/mol):
[tex]\[ \text{Mass of air leaving the room} = \Delta n \times \text{Molar mass} \][/tex]
Let's plug in the given values and calculate the mass of air leaving the room.
First, let's calculate the number of moles of air in the room before and after the temperature change:
[tex]\[ n_1 = \frac{P \times V}{R \times T_1} \][/tex]
[tex]\[ n_1 = \frac{126600 \times 123}{8.314 \times (22.2 + 273.15)} \][/tex]
[tex]\[ n_1 \approx 757.02 \, \text{moles} \][/tex]
[tex]\[ n_2 = \frac{126600 \times 123}{8.314 \times (31 + 273.15)} \][/tex]
[tex]\[ n_2 \approx 722.59 \, \text{moles} \][/tex]
Now, let's find the difference in the number of moles:
[tex]\[ \Delta n = n_1 - n_2 \][/tex]
[tex]\[ \Delta n \approx 757.02 - 722.59 \][/tex]
[tex]\[ \Delta n \approx 34.43 \, \text{moles} \][/tex]
Finally, let's calculate the mass of air leaving the room:
[tex]\[ \text{Mass of air leaving the room} = \Delta n \times \text{Molar mass} \][/tex]
[tex]\[ \text{Mass of air leaving the room} = 34.43 \times 38.1 \][/tex]
[tex]\[ \text{Mass of air leaving the room} \approx 1311.783 \, \text{grams} \][/tex]
So, the mass of air leaving the room is approximately 1311.783 grams.
