# Compression of a Jaguar XK8 Cylinder

1. Jun 18, 2008

### Number1Ballar

1. The problem statement, all variables and given/known data
A Jaguar XK8 convertible has an eight-cylinder engine. At the beginning of its compression stroke, one of the cylinders contains 499 cm^3 of air at atmospheric pressure (1.01 x 10^5 Pa) and a temperature of 27.0 degrees celsius. At the end of the stroke, the air has been compressed to a volume of 46.2 cm^3 and the gauge pressure has increased to 2.72 x 10^6 Pa.

2. Relevant equations
(p1)(V1)/T1 = (p2)(V2)/T2
where p = pressure, V = volume, T = temperature

3. The attempt at a solution
What I tried was:
((1.01 x 10^5)(0.0499))/300 = ((2.72 x 10^6)(4.62 x 10^-4))/T2

I converted the 499cm cubed and the 46.2cm cubed to meters cubed, and also for T1 I did 273(kelvins) + 27

There was a hint given that the 'gauge pressure is the difference between the absolute pressure and the atmospheric pressure. Thus, if you measure a gauge pressure pg, then the absolute pressure p is given by p= pa + pg, where pa is the atmospheric pressure.' So I was thinking for p2 to use ((1.01 x 10^5)+(2.72 x 10^6)) but still did not get the right answer.

The answer to this question is 503 degrees Celsius but I just can't figure out what number to use for p2.

2. Jun 18, 2008

### Dick

You don't have to convert the cm^3 to m^3. If they are both in the same units, just use them as they are. The units will cancel out. The pressure goes up by a factor of about 20 and the volume drops by a factor of about 10. The temperature should about double. You are making a mistake in converting the volumes. 499cm^3=4.99*10^(-4)m^3. 46.2cm^3=4.62*10^(-5)m^3. And, yes, use the gauge pressure thing to make fine adjustments.

Last edited: Jun 18, 2008
3. Jun 18, 2008

### Number1Ballar

Oh okay thank you for the clarification! I was able to get the answer :)