An automobile tire at 22oC with an internal volume of 20.0L is filled with air to a total pressure of 30. psi(pounds per square inch). [1 atm = 14.696 lb/in2].
What is the amount in moles of air in the tire?
If the air is mostly nitrogen, how many grams of it would be in the tire?
How many pounds of it would be in the tire?
To find the amount in moles of air in the tire, we will use the ideal gas equation:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.
First, let's convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 22 + 273.15
T(K) = 295.15 K
Next, let's convert the pressure from psi to atm:
P(atm) = P(psi) / 14.696
P(atm) = 30 / 14.696
P(atm) ≈ 2.04 atm
Now we have all the necessary values to calculate the number of moles:
PV = nRT
n = PV / RT
n = (2.04 atm) * (20.0 L) / [(0.0821 L·atm/mol·K) * (295.15 K)]
n ≈ 1.37 moles of air
To find the number of grams of nitrogen in the tire, we need to know the composition of air. We'll assume that air is 78% nitrogen (N2).
Mass of nitrogen = (78/100) * molar mass of N2 * number of moles
Molar mass of N2 = 2 * atomic mass of nitrogen ≈ 2 * 14.01 g/mol ≈ 28.02 g/mol
Mass of nitrogen = (0.78) * (28.02 g/mol) * (1.37 moles)
Mass of nitrogen ≈ 30.40 g
Finally, to convert grams of nitrogen to pounds, we'll divide the mass by the conversion factor of 453.59 grams per pound:
Mass of nitrogen (in pounds) ≈ 30.40 g / 453.59 g/lb
Mass of nitrogen (in pounds) ≈ 0.067 pounds
Therefore, there are approximately 1.37 moles of air, 30.40 grams of nitrogen, and 0.067 pounds of nitrogen in the tire.