Question

Ideal Gas Law I:

Part A
An 8.50-L tire contains 0.552 mol of gas at a temperature of 305 K. What is the pressure, in atm and in psi, of the gas in the tire? (1 atm is approximately equal to 14.7 psi.)
Express the pressures in atmospheres and pounds per square inch to three significant figures separated by a comma.

Answer 1

pV=nRT; p=nRT/V

p=0.552Mol*8.314J/MolᵒK*305K/0.0085m³=164675.887Pa

p=1.625atm

1.625atm*14.7psi/1atm=23.888psi