Calculate Tire Pressure After Compression & Heating

[trah22]

Homework Statement

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, what's tire pressure?

Homework Equations

pv=nrt so p=nrt/v

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 that's right(i also posted this question in the advanced physics forum by too i wasnt sure where to put it)

 

[Doc Al]

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?)

 

[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 don't see exactly how to relate them according to what u said...

 

[Doc Al]

...is pv/t=constant because pv=nrt=Nkt?

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?

 

[trah22]

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

 

[Doc Al]

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

 

[trah22]

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

 

[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%, what's 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...

 

[Doc Al]

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

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

(P2)V2/(313)=((P3(V2(.2)/(358)
P3=358(p2)(v2)/313(v2*.2)
=358p2(v2)/62.6v2
=5.718(P2)

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

but according to the question i need to put it in pascals...

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.

 

[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 here's 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 that's way to large of a value, did i plug in the worng value for the pressure.. i just used the coversion value for pascals...

 

[Doc Al]

for t3 i just took the 85 degrees celcius and converted to 358K

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%.

 

[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

 

[Doc Al]

oh so it should have been v2(.98)

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

 

[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..

 

[Doc Al]

Now you've got it.

 

[trah22]

heh thanks for all the help again doc