# Ideal Gas Equation problem

1. Aug 27, 2015

### Luchekv

1. The problem statement, all variables and given/known data

A tire is checked before a road trip and the gauge pressure reads 220 kPa (gauge) - State 1
The same tire is checked after the trip and the gauge pressure reads 240 kPa (gauge) - State 2
The temperature on the day is 25 degrees Celsius and the pressure is 101.325 kPa (absolute)
The tire has an approximate inner volume of 25L and can be assumed to be rigid
Gas Constant R for air = 0.2870

A.) Calculate the abs pressure in the tire at state-1.
B.) Using the ideal gas equation, find the mass in grams of the air inside a single tire at state-1.
C.) Using the ideal gas equation, find the temperature of the air inside a single tire at state-2
D.) If having attained state-2, the driver wants to bleed-off some ‘hot’ air from each tire in order to restore the (gauge) pressure to 220kPa, calculate the remaining mass of air inside a single tire after some air is removed

2. Relevant equations
- PV = mRT
- P1*V1/ T1 = P2*V2/ T2
- Pabs=Pg+Patm

3. The attempt at a solution
A.) Pabs=Pg+Patm

P1 = 220kPa + 101.325kPa = 321.325kPa
P2 = 240kPa + 101.325kPa = 341.325kPa

B.) PV=mRT ∴ m= PV/RT
321.325kPa * 25L / 0.2870 *(25+273) = 939261.11 grams

C.) P1*V1/ T1 = P2*V2/ T2
Since V = cosntant the equation is then P1/ T1 = P2/ T2
T2= (P2/P1) * T1
T2
= (341.325/321.325) * (25+273) = 43.55 Degrees Celsius

D.) PV=mRT ∴ m= PV/RT
321.325kPa * 25L / 0.2870 *(43.55+273) = 88421.99 grams

2. Aug 27, 2015

### Staff: Mentor

Units?

What does part A ask for?

939 kg??? Hope you don't need to change a tire!

3. Aug 27, 2015

### Luchekv

I decided to do P2 while I was at it since I would have needed it later on.

Sorry, 0.2870 Kj/kg.Kelvin **

As for the the 939kg, I wasn't sure if I was meant to leave 321.325kPa as just that "321.325" or 321.325*10^3..
Otherwise it would be 93.92 grams
Thank you again :)

4. Aug 27, 2015

### Staff: Mentor

The mistake was not in the pressure. That's one of the reasons I asked you to specify the units: make sure they are consistent.

5. Aug 27, 2015

### Luchekv

Unfortunately I'm terrible when it comes to differing units.. I didn't see any problem with the R constant...however I did forget to convert the 25L to m^3. Was that it?

6. Aug 27, 2015

### Staff: Mentor

Yes.

7. Aug 27, 2015

### Luchekv

Thank you! :)

8. Aug 27, 2015

### SteamKing

Staff Emeritus
Or it could be that pressure needed to have units of kPa instead of just Pa.

The tricky thing about R is that it can come in a variety of units, which is why it is important to establish these before doing the arithmetic.