An apartment has a living room whose dimensions are 2.5 m 4.8 m 5.9 m. Assume that the air in the room is composed of 79% nitrogen (N2) and 21% oxygen (O2). At a temperature of 27°C and a pressure of 1.01 105 Pa, what is the mass (in grams) of the air?
You don't say but I think you must be quoting percent volume.
Find volume of room, I suggest, in cc and change to liters.
Change liters at conditions listed to standard conditions. Standard conditions are 101 325 Pa for pressure and 273 K for temperature. Use P1V1/T1 = P2V2/T2
corrected volume x 0.79 = volume N2 @ STP.
corrected volume x 0.21 = volume O2@ STP.
1 mol of a gas occupies 22.4 L @ STP; therefore,
volume N2/22.4 = mols N2
mols N2 x molar mass N2 = grams N2.
volume O2/22.4 = mols O2
mols O2 x molar mass O2 = grams O2.
grams N2 + grams O2 = mass of air in the room. Post your work if you need more help.
To calculate the mass of the air in the living room, we first need to determine the volume of the room. The volume can be calculated by multiplying its dimensions: length ̶ 2.5 m, width ̶ 4.8 m, and height ̶ 5.9 m.
Volume of the room = length x width x height
= 2.5 m x 4.8 m x 5.9 m
= 70.8 m^3
The next step is to use the Ideal Gas Law to find the number of moles of air in the room. The Ideal Gas Law states:
PV = nRT
Where:
P = Pressure (in Pa)
V = Volume (in m^3)
n = Number of moles
R = Universal Gas Constant (8.314 J/(mol·K))
T = Temperature (in Kelvin)
First, we need to convert the temperature from Celsius to Kelvin by adding 273.15:
Temperature in Kelvin = 27°C + 273.15
= 300.15 K
Now, rearrange the Ideal Gas Law equation to solve for n (number of moles):
n = PV / (RT)
Given:
P = 1.01 x 10^5 Pa
V = 70.8 m^3
R = 8.314 J/(mol·K)
T = 300.15 K
Substitute the values into the equation:
n = (1.01 x 10^5 Pa) x (70.8 m^3) / (8.314 J/(mol·K) x 300.15 K)
Simplify the expression:
n = 471620.241 / 2494.06841
n ≈ 189.156 mol
Now that we have the number of moles of air in the room, we can calculate the mass of the air.
To do this, we will use the molar mass of air, which is the weighted average of the molar masses of nitrogen (N2) and oxygen (O2). Nitrogen has a molar mass of 28.0134 g/mol, while oxygen has a molar mass of 31.9988 g/mol.
To find the mass of nitrogen, we multiply the number of moles by the molar mass of nitrogen:
Mass of nitrogen = (Number of moles of nitrogen) x (Molar mass of nitrogen)
Mass of nitrogen = (0.79) x (189.156 mol) x (28.0134 g/mol)
To find the mass of oxygen, we multiply the number of moles by the molar mass of oxygen:
Mass of oxygen = (Number of moles of oxygen) x (Molar mass of oxygen)
Mass of oxygen = (0.21) x (189.156 mol) x (31.9988 g/mol)
Finally, we add the mass of nitrogen and the mass of oxygen to get the total mass of the air in the room:
Total mass of air = Mass of nitrogen + Mass of oxygen
To summarize, first, calculate the volume of the room by multiplying its dimensions. Then, use the Ideal Gas Law to find the number of moles of air. Next, calculate the mass of nitrogen and oxygen separately by multiplying the number of moles with their respective molar masses. Finally, add the masses of nitrogen and oxygen to get the total mass of the air in grams.