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

In summary, solving an ideal gas equation problem with changing tire pressures involves applying the ideal gas law, PV=nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. To solve for an unknown variable, rearrange the equation and plug in the given values. When dealing with tire pressures, make sure to convert from units of atm to kPa. Additionally, use the combined gas law, P1V1/T1 = P2V2/T2, to solve for changes in pressure or temperature when other variables are held constant. Finally, remember to use proper units and significant figures in the final answer.
  • #1
Luchekv
66
1
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
- Pabs=Pg+Patm[/B]

The Attempt at a Solution


A.) Pabs=Pg+Patm[/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
 
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  • #2
Luchekv said:
Gas Constant R for air = 0.2870
Units?

Luchekv said:

The Attempt at a Solution


A.) Pabs=Pg+Patm[/B]
P1 = 220kPa + 101.325kPa = 321.325kPa
P2 = 240kPa + 101.325kPa = 341.325kPa
What does part A ask for?

Luchekv said:
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!
 
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  • #3
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
Luchekv said:
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.
 
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  • #5
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
Luchekv said:
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.
 
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  • #7
Thank you! :)
 
  • #8
Luchekv said:
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
 

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

1. What is the Ideal Gas Equation?

The Ideal Gas Equation, also known as the General Gas Equation, is a mathematical formula that describes the behavior of an ideal gas. It relates the pressure, volume, temperature, and amount of a gas to each other using the variables P, V, T, and n, respectively.

2. What are the units used in the Ideal Gas Equation?

The units used in the Ideal Gas Equation depend on the units used for pressure, volume, and temperature. The most common units are atmospheres (atm) for pressure, liters (L) for volume, and Kelvin (K) for temperature. The amount of gas, n, is typically measured in moles (mol).

3. How is the Ideal Gas Equation used in real-life scenarios?

The Ideal Gas Equation is used in many real-life scenarios, such as predicting the behavior of gases in chemical reactions, designing and operating refrigeration systems, and understanding the behavior of air in weather patterns. It is also used in the study of thermodynamics and in the development of new materials and processes.

4. What are the assumptions made in the Ideal Gas Equation?

The Ideal Gas Equation makes several assumptions about the behavior of an ideal gas, including that the gas particles are point masses with no volume, that they do not interact with each other, and that the volume of the gas is much larger than the volume of the particles. These assumptions allow for simplified calculations and predictions, but may not accurately reflect the behavior of real gases.

5. What are some common applications of the Ideal Gas Equation problem?

The Ideal Gas Equation can be used to solve a variety of problems, such as calculating the pressure, volume, or temperature of a gas under different conditions, determining the amount of a gas needed for a reaction, or predicting the behavior of a gas in a closed container. It is also commonly used in chemistry and physics experiments, as well as in industrial processes and engineering designs.

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