Final pressure homework

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  • #1
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A tire is filled with air at 10°C to a gauge pressure of 220 kPa. If the tire reaches a temperature of 30°C, what fraction of the original air must be removed if the original pressure of 220 kPa is to be maintained?

I tried finding the pressure at 30°C assuming volume stays the same using P1/T1=P2/T2 and got nowhere after that. Then I tried setting P to be equal and use the resulting formula of T1/V1=T2/V2 and got a percentage greater than 100%...
 

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  • #2
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First, get the final pressure (use all absolute terms) keeping volume constant. You are right with your first step.

Now keep the temperature constant and express the volume at initial pressure as a sum of initial volume and some increment. Now it is simple arithmetic.
 
  • #3
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I don't understand why you keep the temperature constant...can you explain some more?
 
  • #4
lurflurf
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physicsss said:
A tire is filled with air at 10°C to a gauge pressure of 220 kPa. If the tire reaches a temperature of 30°C, what fraction of the original air must be removed if the original pressure of 220 kPa is to be maintained?

I tried finding the pressure at 30°C assuming volume stays the same using P1/T1=P2/T2 and got nowhere after that. Then I tried setting P to be equal and use the resulting formula of T1/V1=T2/V2 and got a percentage greater than 100%...
So we have
n1T1=n2T2
so
n2/n1=T1/T2
n= amount of substance
T= temperature (absolute)
T1=(273.15+10)k=283.15k
T2=(273.15+30)k=303.15k
we desire to find
(n1-n2)/n1=1-n2/n1
fraction of air removed=1-n2/n1=1-T1/T2=1-(283.15k)/(303.15k)=6.6%
The general form for the ideal gas assumption for two sets of conditions is
(P1V1)/(n1T1)=(P2V2)/(n2T2)
P=pressure (absolute)
V=volume
n= amount of substance
T= temperature (absolute)
any that do not change may be droped
here pressure and volume we constant giving
n1T1=n2T2
 
  • #5
231
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Because you wanted to know the amount of air to be removed to obtain the initial pressure and at 30C temperature. Alternately, you can use the procedure given by lurflurf.
 

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