Could you help me to solve this please and just explain the procedure...

1. An automobile tire is inflated to 32 psig pressure at 50 degrees Farenheit. After driven the temperature rise to 75 farenheit. Determine the final gage pressure assuming the volume is constant. Answer: 34.29 psig

2. If 100 ft^3 of atmospheric air at zero Farenheit temperature are compressed to a volume of 1 ft^3 at a temperature of 200 deg Farenheit, what will be the pressure of the air in psi? Answer:2109 psia

pV = nrT (ideal gas law) this should get you started


then make sure you convert degrees F to Celsius.

since V= constant, and n=1,

p=r(t2-t1) where r=rydberg constant.


easy.


p= 8.314472 x (23.8888889 - 20.0)






You can try to work out the second one yourself before asking any more questions.

Sir where did you get the T2 which is 20 because 50 degrees Farenheit is equal to approximately 10 Celcius...

He shouldn't have said Celsius, He should have said Rankine or Kelvin, with a zero at the Absolute Zero, where the ideal gas volume is also zero.

http://en.wikipedia.org/wiki/Rankine_scale

Likewise gauge pressures must be converted to absolute pressures.

So the though process for the first problem looks like this:

PV = NRT So...
R = PV/NT = P₁V₁/N₁T₁ = P₂V₂/N₂T₂
V₁ = V₂ (given)
N₁ = N₂ (because no gas enters or exits)
So
P₁/T₁ = P₂/T₂ (for constant V and N given the ideal gas law)
So P₂ = P₁T₂/T₁
And P(absolute) = p(gauge) + P(atmospheric pressure), and T(Absolute) = t(℉) + 459.67°R
So we get
p₂(gauge) + P(atmospheric pressure) = (p₁(gauge) + P(atmospheric pressure)) (t₂(℉) + 459.67°R)/(t₁(℉) + 459.67°R)
or
p₂(gauge) = (p₁(gauge) + P(atmospheric pressure)) (t₂(℉) + 459.67°R)/(t₁(℉) + 459.67°R) - P(atmospheric pressure)

So we can look up P(atmospheric pressure) = 14.7 psi and calculate this.

So the first calculation (expressed in psi gauge but using intermediates of absolute pressure in psi and absolute temperature in Rankine) looks like this:

http://www.google.com/search?q=what+is+(32...%2B459.67)-14.7

As a side note, Google's calculator looks smart, but is not perfect, so this also works, even though it looks misleading because of the silent conversion to absolute temperatures:
http://www.google.com/search?q=what+is+(32...scals+in+psi%3F

Hello Sir, I am a newbie in thermodynamics, I highly appreciate your clear explanation...thanks a lot