TICKET 1-23 1. Find the mass of air in a room of 25m^2 area and a height of 3.2 m Assume that the tem- perature in the room is 22^circ C and the barometric pressure is B=986.5hPa

To find the mass of air in the room, we can use the ideal gas law equation:<br /><br />\[ PV = nRT \]<br /><br />Where:<br />- \( P \) is the pressure of the gas (in Pa)<br />- \( V \) is the volume of the gas (in m³)<br />- \( n \) is the number of moles of the gas<br />- \( R \) is the universal gas constant (8.314 J/(mol·K))<br />- \( T \) is the temperature of the gas (in K)<br /><br />First, we need to convert the given values to appropriate units:<br /><br />1. **Temperature Conversion:**<br /> \[ T = 22^{\circ}C + 273.15 = 295.15 \, K \]<br /><br />2. **Pressure Conversion:**<br /> Given barometric pressure \( B = 986.5 \, hPa \) can be converted to Pa:<br /> \[ P = 986.5 \, hPa \times 100 \, Pa/hPa = 98650 \, Pa \]<br /><br />3. **Volume Calculation:**<br /> The volume of the room is:<br /> \[ V = \text{Area} \times \text{Height} = 25 \, m^2 \times 3.2 \, m = 80 \, m^3 \]<br /><br />Now, we can use the ideal gas law to find the number of moles of air in the room:<br /><br />\[ n = \frac{PV}{RT} \]<br /><br />Substitute the known values into the equation:<br /><br />\[ n = \frac{98650 \, Pa \times 80 \, m^3}{8.314 \, J/(mol \cdot K) \times 295.15 \, K} \]<br /><br />\[ n = \frac{7882000 \, Pa \cdot m^3}{2458.7 \, J/(mol \cdot K)} \]<br /><br />\[ n \approx 3211.6 \, mol \]<br /><br />Finally, to find the mass of the air, we multiply the number of moles by the molar mass of air (approximately 28.97 g/mol):<br /><br />\[ \text{Mass} = n \times \text{Molar Mass} \]<br /><br />\[ \text{Mass} = 3211.6 \, mol \times 28.97 \, g/mol \]<br /><br />\[ \text{Mass} \approx 92600 \, g \]<br /><br />\[ \text{Mass} \approx 92.6 \, kg \]<br /><br />Therefore, the mass of air in the room is approximately 92.6 kg.