A bicycle pump inflates a tire whose volume is 577 mL until the internal pressure is 627 kPa at a temperature of 21.7°C. How many moles of air does the tire contain?
PV = nRT
If you use P in kPa, R is 8.314
or if you use P in atm (627 kPa/101.325) then R is 0.08206. Solve for n.
To find the number of moles of air in the tire, we can use the ideal gas law equation:
PV = nRT
Here,
P is the pressure (627 kPa),
V is the volume in liters (577 mL = 0.577 L),
n is the number of moles of air,
R is the ideal gas constant (8.314 J/(mol·K)),
T is the temperature in kelvin (21.7°C + 273.15 = 294.85 K).
Rearranging the equation to solve for n, we get:
n = PV / RT
Let's plug in the values:
n = (627 kPa * 0.577 L) / (8.314 J/(mol·K) * 294.85 K)
Now let's solve it step by step:
1. Convert kilopascals (kPa) to pascals (Pa):
627 kPa = 627,000 Pa
2. Calculate:
n = (627,000 Pa * 0.577 L) / (8.314 J/(mol·K) * 294.85 K)
3. Simplify:
n = (362,079,000 J·L) / (2447.498 J·mol^(-1)·K^(-1))
4. Cancel out the units:
n = (362,079,000 J·L) / (2447.498 J·mol^(-1)·K^(-1)) = 148,056 mol
So, the tire contains approximately 148,056 moles of air.