# Pressure problem

1. Jul 14, 2007

### trah22

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

automobile tire is inflated with air originally at 10 degrees celcius and at normal atmosphere pressure. During the process the air is compressed to 28% of its original volume and the tempreture is increased to 40 degrees celcius, whats tire pressure?

2. Relevant equations

pv=nrt so p=nrt/v

3. The attempt at a solution
p=(not sure what to use for n)(.0821)(30K)/(72)

i subracted 28 from 100 to get 72 for volume not sure if thats right(i also posted this question in the advanced physics forum by too i wasnt sure where to put it)

Last edited: Jul 14, 2007
2. Jul 14, 2007

### Staff: Mentor

Realize that PV/T = constant: use ratios. If you call the original volume V, what is the final volume? What are the initial and final temperatures? (What temperature scale must you use?)

3. Jul 14, 2007

### trah22

the temp has to be in kelvin, according to my notes for equation of state, so the intial temp was 283 K and the final is 313 hence my 30K value in my attempted solution.... um is pv/t=constant because pv=nrt=Nkt? I know volume and temp are related(if temp increases volume increases) but i dont see exactly how to relate them according to what u said...

4. Jul 14, 2007

### Staff: Mentor

Yes.

Try this:

$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$

Fill in the values as best you can. Hint: How would you write $V_2$ in terms of $V_1$?

5. Jul 14, 2007

### trah22

well, P1(v1)/283K=P2(v1*.28)/313..
arghh jeez as u can tell physics is not my fortay, im a bio-chem major

6. Jul 14, 2007

### Staff: Mentor

Good. Now just solve for P2 in terms of P1 and you're done.

7. Jul 14, 2007

### trah22

P2=(313(P1)(V1))/((v1(.28)(283))
=313(P1)(v1)/(79.24(V1))
=3.95(p1)
did calculate it out correctly?

Last edited: Jul 14, 2007
8. Jul 14, 2007

### trah22

theres 1 more part to this question, after the car is driven at a high speed, the tire air temp rose to 85 degrees celcius and the interior volume of the tire increased by 2%, whats the new tire pressure in pascals?

(P2)V2/(313)=((P3(V2(.2)/(358)
P3=358(p2)(v2)/313(v2*.2)
=358p2(v2)/62.6v2
=5.718(P2)
but according to the question i need to put it in pascals...

9. Jul 15, 2007

### Staff: Mentor

Looks good. Note that you are given P1--that's atmospheric pressure. So you can express your answer in terms of Pascals.

Where did you get T3 = 358? And if the volume increases by 2%, what is V3 in terms of V2?

You are finding P3 in terms of P2, but you know P2 in terms of P1--and thus Pascals--from the first part of the problem.

10. Jul 15, 2007

### trah22

alright, 1atm=1.013x10^5 pascals so P2=4.0x10^5 pascals,
for t3 i just took the 85 degrees celcius and converted to 358K

alright heres my go at the problem again in terms of volume and pressure,

4x10^5Pa(v2)/313K=P3(V2(.02)/358
P3=358(4x10^5)(v2)/313(v2(.02)
p3=1.43x10^8/6.26
p3=2.28x10^7Pa

i had to have messed up somewhere thats way to large of a value, did i plug in the worng value for the pressure.. i just used the coversion value for pascals...

11. Jul 15, 2007

### Staff: Mentor

OK, good. But V3 does not equal V2(0.02). That's way off. The volume increases by 2%--you have it decreasing by 98%.

12. Jul 15, 2007

### trah22

oh so it should have been v2(.98)

so P3=4x10^5Pa(358)(v2)/313(v2(.98)
=4.668x10^5Pa
does this large of a value for pressure seem correct but anyhow thanks for all the help,i appreciate it

13. Jul 15, 2007

### Staff: Mentor

No, but you're getting closer. This time you have the volume decreasing by 2%.

14. Jul 15, 2007

### trah22

interior volume increases by 2%, so the original v2 is 100% plus 2% increase so v2(1.02)..... ud just add the 2% since it just increases then..

15. Jul 15, 2007

### Staff: Mentor

Now you've got it.

16. Jul 15, 2007

### trah22

heh thanks for all the help again doc