Solving Gas Law Question: Tire Pressure at 14°C & 2.2atm

  • Thread starter Thread starter iluvu
  • Start date Start date
  • Tags Tags
    Gas Gas law Law
AI Thread Summary
The discussion revolves around calculating the temperature of a car tire after its pressure increases from 2.0 atm to 2.2 atm while the external temperature remains at 14°C. Participants reference gas laws, including Charles's Law and Boyle's Law, to approach the problem. There is some confusion regarding the use of atmospheric pressure versus kilopascals, but it is clarified that the unit conversion is unnecessary for solving the question. The key focus is on understanding how the pressure change relates to the temperature change in the tire. Ultimately, the discussion aims to apply the combined gas law to find the unknown temperature in the tire.
iluvu
Messages
2
Reaction score
0
The question given were:

The tire pressure for a car that has not been driven is 2.0atm when sitting outside at 14 degrees celsius. After the car was on the highway the temperature of the tire increased even though the air temperature remained at 14 degrees celsius. The tire pressure increased to 2.2atm. Calculate the temperature of the air in the tire in degrees celsius.

The gas laws that the teacher ever told us is

charles law: V1/T1=V2/T2
Bogle's law is P1V1=P2/V2 and
Combine law, which is V1*P1/T1=V2*P2/T2

But in the question, there is atm...which stands for Atmosphere...I know that Atomosphere is a unit of pressure, but isn't the unit for pressure (kPa)?

I'm so stuck...=( someone please help me...
 
Physics news on Phys.org
1 atm is approximately 1.013 \times 10^5 Pa. But you do not need to perform this conversion to solve the problem because this factor will eventually cancel off.

What quantities remain the same before and after the car was driven?
 
Last edited:
1 kPa is 1000 Pa
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
I don't get how to argue it. i can prove: evolution is the ability to adapt, whether it's progression or regression from some point of view, so if evolution is not constant then animal generations couldn`t stay alive for a big amount of time because when climate is changing this generations die. but they dont. so evolution is constant. but its not an argument, right? how to fing arguments when i only prove it.. analytically, i guess it called that (this is indirectly related to biology, im...
Back
Top