How to Solve an Ideal Gas Equation Problem with Changing Tire Pressures?

[Luchekv]

Hey guys, I just want to make sure I went about this the right way...your input would be greatly appreciated.
Thank you in advance.

1. Homework Statement
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

It asks for:
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

Homework Equations

- PV = mRT
- P₁*V₁/ T₁ = P₂*V₂/ T₂
- P_abs=P_g+P_atm[/B]

The Attempt at a Solution

A.) P_abs=P_g+P_atm[/B]
P₁ = 220kPa + 101.325kPa = 321.325kPa
P₂ = 240kPa + 101.325kPa = 341.325kPaB.) 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

 

[DrClaude]

Gas Constant R for air = 0.2870

Units?

The Attempt at a Solution

A.) P_abs=P_g+P_atm[/B]
P₁ = 220kPa + 101.325kPa = 321.325kPa
P₂ = 240kPa + 101.325kPa = 341.325kPa

What does part A ask for?

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

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

 

[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 :)

 

[DrClaude]

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 :)

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.

 

[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?

 

[DrClaude]

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?

Yes.

 

[Luchekv]

Thank you! :)

 

[SteamKing]

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?

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.

Take a look at R in this article:

https://en.wikipedia.org/wiki/Gas_constant

R can have the same numerical value (8.314), but the units could be L⋅kPa⋅K^-1⋅mol^-1 or m³⋅Pa⋅K^-1⋅mol^-1