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?
g
V = 2.5 * 4.8 * 5.9 = 70.8 m^3
N is 14 so N2 is 28 grams/mole
O is 16 so O2 is 32 grams/mole
so a mole of "air" has mass of .79*28 + .21*32 = 28.8 grams/mole
assume air is ideal gas
P V = n R T
1.01 * 10^5 * 70.8 = n * 8.3145 *(273+27)
n = 3037 moles
3037* 28.8 = 87468 grams
my teacher says this is the incorrect answer. what could be wrong
Well, let's not take this question lightly! To calculate the mass of air in the room, we'll start by finding the number of moles of each gas using the ideal gas law equation:
PV = nRT
Where:
P = pressure (in pascals)
V = volume (in cubic meters)
n = number of moles
R = ideal gas constant (8.314 J/mol*K)
T = temperature (in Kelvin)
First, let's convert the temperature from Celsius to Kelvin:
27°C + 273.15 = 300.15 K
Now let's calculate the number of moles of each gas:
For nitrogen (N₂):
P = 1.01 x 10⁵ Pa
V = 2.5 m * 4.8 m * 5.9 m = 70.8 m³
R = 8.314 J/mol*K
T = 300.15 K
Using the ideal gas law equation: PV = nRT, we can rearrange it to solve for n:
n(N₂) = PV / RT
n(N₂) = (1.01 x 10⁵ Pa)(70.8 m³) / (8.314 J/mol*K)(300.15 K)
Now let's do the calculations. But wait for it... Remember to convert the units!
n(N₂) ≈ [(1.01 x 10⁵ N/m²)(70.8 m³)] / [(8.314 kg·m²·s⁻²·K⁻¹·mol⁻¹)(300.15 K)]
The units cancel out, but let me use this opportunity to say that canceling out units is a skill that would come in handy when canceling out awkward conversations or canceling out plans with friends you don't want to hang out with. Anyway, back to the calculations.
n(N₂) ≈ 854.58138 mol
And now, let's move on to oxygen (O₂). We'll follow the same steps:
For oxygen (O₂):
P = 1.01 x 10⁵ Pa
V = 2.5 m * 4.8 m * 5.9 m = 70.8 m³
R = 8.314 J/mol*K
T = 300.15 K
n(O₂) ≈ [(1.01 x 10⁵ N/m²)(70.8 m³)] / [(8.314 kg·m²·s⁻²·K⁻¹·mol⁻¹)(300.15 K)]
Calculating, we find:
n(O₂) ≈ 227.76496 mol
Now that we have the number of moles of each gas, time for the final step: calculating the mass. Each mole of nitrogen (N₂) has a molar mass of approximately 28 g/mol, and each mole of oxygen (O₂) has a molar mass of approximately 32 g/mol.
So, the mass of nitrogen in the room is:
Mass(N₂) = n(N₂) x m(N₂)
Mass(N₂) ≈ 854.58138 mol × 28 g/mol
After the calculations, we find:
Mass(N₂) ≈ 23,922.907 g
And the mass of oxygen is:
Mass(O₂) = n(O₂) x m(O₂)
Mass(O₂) ≈ 227.76496 mol × 32 g/mol
After more calculations, we find:
Mass(O₂) ≈ 7,288.31872 g
To find the total mass of air, we simply add the masses of nitrogen and oxygen:
Total mass = Mass(N₂) + Mass(O₂)
Total mass ≈ 23,922.907 g + 7,288.31872 g
Doing the math, we find:
Total mass ≈ 31,211.22572 g
So the mass of the air in the room is approximately 31,211.23 grams. Keep in mind that this is just the mass of the air, not the weight of responsibility that comes with living in an apartment.
To find the mass of the air in the room, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure in pascals = 1.01 x 10^5 Pa
V = volume in cubic meters = 2.5 m x 4.8 m x 5.9 m = 71.04 m^3
n = number of moles of air
R = ideal gas constant = 8.314 J/(mol·K)
T = temperature in kelvin = 27°C + 273.15 = 300.15 K
We can rearrange the equation to solve for n:
n = PV / RT
n = (1.01 x 10^5 Pa) x (71.04 m^3) / (8.314 J/(mol·K) x 300.15 K)
Now, let's convert the result into grams:
1 mole of air has a molar mass of approximately 28.97 g, which is the average of the molar masses of nitrogen and oxygen.
molar mass of air = 28.97 g/mol
So, the mass of the air in grams can be calculated by multiplying the number of moles by the molar mass:
mass of air = n x molar mass of air
Substitute the values:
mass of air = [(1.01 x 10^5 Pa) x (71.04 m^3) / (8.314 J/(mol·K) x 300.15 K)] x 28.97 g/mol
Calculate this expression to find the mass of the air in grams.
To calculate the mass of the air in the living room, we need to use the ideal gas law equation:
PV = nRT
Where:
P = Pressure (in Pa)
V = Volume (in m^3)
n = Number of moles of gas
R = Ideal gas constant (8.314 J/(mol·K))
T = Temperature (in Kelvin)
First, let's convert the given temperature from °C to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 27 + 273.15
T(K) = 300.15 K
Next, we need to calculate the volume of the living room:
V = length × width × height
V = 2.5 m × 4.8 m × 5.9 m
V = 71.28 m^3
Now, we can plug in the values into the ideal gas law equation and solve for the number of moles of gas (n):
PV = nRT
n = (PV) / (RT)
n = (1.01 × 10^5 Pa) × (71.28 m^3) / ((8.314 J/(mol·K)) × (300.15 K))
n ≈ 3055.52 mol
Since the air in the living room is composed of 79% nitrogen and 21% oxygen, we can calculate the mass of each component separately:
Mass of nitrogen (N2):
m(N2) = n(N2) × M(N2)
where:
n(N2) = number of moles of nitrogen
M(N2) = molar mass of nitrogen (28.02 g/mol)
n(N2) = 3055.52 mol × 0.79
n(N2) ≈ 2413.98 mol
m(N2) = 2413.98 mol × 28.02 g/mol
m(N2) ≈ 67618.68 g
Mass of oxygen (O2):
m(O2) = n(O2) × M(O2)
where:
n(O2) = number of moles of oxygen
M(O2) = molar mass of oxygen (32 g/mol)
n(O2) = 3055.52 mol × 0.21
n(O2) ≈ 641.66 mol
m(O2) = 641.66 mol × 32 g/mol
m(O2) ≈ 20533.12 g
Finally, we can calculate the total mass of the air by adding the masses of nitrogen and oxygen:
Total mass of air = m(N2) + m(O2)
Total mass of air ≈ 67618.68 g + 20533.12 g
Total mass of air ≈ 88151.8 g
Therefore, the mass of the air in the living room is approximately 88151.8 grams.