Questions: Imagine you are sitting in a room with a volume of 27000 liters and a temperature of 72°F. The total air pressure is 785 mm Hg, and the partial pressure from oxygen is 21% of the total pressure. How many grams of oxygen are in the room? amount of oxygen: grams Enter numeric value

Solution Steps

The volume of the room is already given in liters, so no conversion is necessary. The volume \( V \) is: \[ V = 27000 \, \text{liters} \]

First, convert the temperature from Fahrenheit to Celsius: \[ T_{\text{C}} = \frac{5}{9} (T_{\text{F}} - 32) \] \[ T_{\text{C}} = \frac{5}{9} (72 - 32) = \frac{5}{9} \times 40 = 22.2222 \, ^\circ \text{C} \]

Next, convert the temperature from Celsius to Kelvin: \[ T_{\text{K}} = T_{\text{C}} + 273.15 \] \[ T_{\text{K}} = 22.2222 + 273.15 = 295.3722 \, \text{K} \]

The partial pressure of oxygen is 21% of the total pressure: \[ P_{\text{O}_2} = 0.21 \times 785 \, \text{mm Hg} \] \[ P_{\text{O}_2} = 164.85 \, \text{mm Hg} \]

Convert the pressure from mm Hg to atmospheres (1 atm = 760 mm Hg): \[ P_{\text{O}_2} = \frac{164.85}{760} \, \text{atm} \] \[ P_{\text{O}_2} = 0.2167 \, \text{atm} \]

The ideal gas law is given by: \[ PV = nRT \] where:

• \( P \) is the pressure in atmospheres,
• \( V \) is the volume in liters,
• \( n \) is the number of moles,
• \( R \) is the ideal gas constant (\( R = 0.0821 \, \text{L} \cdot \text{atm} / \text{K} \cdot \text{mol} \)),
• \( T \) is the temperature in Kelvin.

Rearrange to solve for \( n \): \[ n = \frac{PV}{RT} \] \[ n = \frac{0.2167 \times 27000}{0.0821 \times 295.3722} \] \[ n = \frac{5850.9}{24.2391} \] \[ n = 241.36 \, \text{moles} \]

The molar mass of oxygen (\( \text{O}_2 \)) is 32.00 g/mol. Therefore, the mass \( m \) in grams is: \[ m = n \times \text{molar mass} \] \[ m = 241.36 \times 32.00 \] \[ m = 7723.52 \, \text{grams} \]

Final Answer