Assume that a bicycle tire with an internral volume of 1.50 liters contains 0.405 mole of air. The tire will burst if the internal pressure is 7.25 atm. What is the temperature of the air in the tire if a blowout occurs?
PV = nRT
You have all except T. Solve for T in Kelvin.
So the equation would be:
PV/nR = T
yes
Just to make sure:
(7.25atm)(1.50 L)/(0.405 mol)(0.0821 L atm/l mol) = T
Final answer: 327 K
looks ok to me.
To find the temperature of the air in the tire when a blowout occurs, we can use the ideal gas law equation:
PV = nRT
Where:
P = Pressure
V = Volume
n = number of moles
R = Ideal gas constant
T = Temperature
We are given:
Pressure (P) = 7.25 atm
Volume (V) = 1.50 liters
Number of moles (n) = 0.405 mole
The ideal gas constant (R) is typically given in units of liter*atm/(mol*K), so we need to make sure all the units are consistent.
First, let's convert the volume from liters to moles by dividing it by the molar volume at STP (standard temperature and pressure). The molar volume at STP is approximately 22.4 L/mol.
Volume (V) = 1.50 L / 22.4 L/mol ≈ 0.067 mol
Now we have:
Pressure (P) = 7.25 atm
Volume (V) = 0.067 mol
Number of moles (n) = 0.405 mole
Let's rearrange the ideal gas law equation to solve for temperature (T):
T = (PV) / (nR)
Substituting the given values:
T = (7.25 atm * 0.067 mol) / (0.405 mol * R)
Now we need to determine the value of the ideal gas constant (R). The ideal gas constant can be expressed as:
R = 0.0821 L*atm/(mol*K)
By substituting the value for R, we get:
T = (7.25 atm * 0.067 mol) / (0.405 mol * 0.0821 L*atm/(mol*K))
T = (0.48675 atm*mol) / (0.0331705 L*atm/(mol*K))
Cancel out the units and calculate:
T ≈ 14.68 K
Therefore, the temperature of the air in the tire when a blowout occurs is approximately 14.68 K.