Calculate Pressure of Air in a Cylinder

AI Thread Summary
To calculate the pressure of air in a cylinder with a diameter of 4cm and height of 12cm when compressed to half its height, one must consider the Ideal Gas Law and the Combined Gas Law, which account for temperature and pressure changes. The method of compression significantly affects the outcome; adiabatic compression results in increased temperature and pressure, while isothermal compression maintains temperature, leading to different pressure calculations. Understanding these principles is crucial for accurate pressure determination. The specific equations for adiabatic and isothermal processes must be applied based on the compression method chosen. Accurate calculations will depend on these factors.
smeating
Messages
1
Reaction score
0
I am starting out with a cylinder that measures 4cm diameter and height is 12cm with the air being under no pressure. I then want to colapse the cylinder down to half of it's original height, pressurizing the cylinder. Can someone tell me what the pressure would be.

Thanks in advance for any help

cheers,
 
Physics news on Phys.org
You may need to consider things like temperature, too - Google the Ideal Gas Law and read about the assumptions made for an Ideal Gas, then search for the Combined Gas Law, which is really probably what you're looking for.

-Kerry
 
KLoux said:
You may need to consider things like temperature, too - Google the Ideal Gas Law and read about the assumptions made for an Ideal Gas, then search for the Combined Gas Law, which is really probably what you're looking for.

-Kerry

This depends on the method of compression. If it is compressed rapidly, (i.e. adiabatically) both the temperature and pressure will increase. If the gas is allowed to return to the temperature it was before compression, the pressure will be different than in the previous example. There are different equations for each process (adiabatic and isothermal).
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top