1. Oct 3, 2008

### jabjab

1. The problem statement, all variables and given/known data
hi,
can anyone help me with this problem?

a motorist equips his automobile tire with a relief type valve so that the pressure inside the tire never exceeds 240 KPA(gage). He starts a trip with a pressure of 200 KPA(gage) and a temp. 23'C in the tires.During the long drive the temp. of air in the tire reaches 83'C each tire contains 0.11kg of air.a) the mass of air escaping each tire.b) the pressure of the tire when temp. return to 23'C

given the answer provide the solution:

2. Relevant equations

3. The attempt at a solution
for a)
i use the formula PV=mRT and derive equation:

m1=P1V1/R(T1)
m2=P2V2/R(T2)

i assume volume is constant because on the problem volume is not given
m1= ((200KJ/m3)(1m3)(1000J/1KJ))/((286.9J/Kg.K)(296K)
=20000/84922.4
=2.35509kg

m2=((240)(1)(1000))/((286.9)(356))
= 240000/102136.4
= 234980kg

m1-m2=m
m= 2.35509kg-2.34980kg
m= 0.00529kg

for b)
i use the formula
P1/T1=P2/T2

P2=P1(T2/T1)
=240KPA(gage)(296/356)
=199 KPA(gage)

2. Oct 3, 2008

### dlgoff

Isn't the equation PV=nRT where n is the number of moles of the gas?

3. Oct 3, 2008

### Andrew Mason

You have to determine the number of moles of air in the tire from the mass of air. Then you have to find the volume of the tire using PV = nRT where R is the gas constant 8.314 Joules/mole K.

What is the number of moles of air in the tire to begin at T = 296 K? Find V (which is the same at all times).

Then just apply PV=nRT.

AM