Ok here is the scenario, I have an empty cup and a piston that forms an airtight seal. The gas in side is just standard air, at 14.7 psi, and 25ºC. If I compress the piston by 50% so the volume the of gas inside is 1/2 what it was before, the pressure would more than twice as high and the temperature would be higher as well.

My question is what formula would I use to calculate the pressure and temperature of the compressed gas?

I have been looking online for a while and cannot find this, but I am probably using the wrong words.

  • $\begingroup$ If someone could do an example that would be very helpful, I cannot seem to understand the formulas below. $\endgroup$ – Robert Short Aug 17 '17 at 0:26

The process you are describing is an adiabatic compression. To work out what happens we use the equation:

$$ PV^\gamma = \text{constant} \tag{1} $$

where $\gamma$ is the ratio of the two specific heats $C_P/C_V$.

In this case if the initial state is $P_1$, $V_1$ and the final state $P_2$, $V_2$ then equation (1) gives us:

$$ P_1V_1^\gamma = P_2V_2^\gamma $$

or with a quick rearrangement:

$$ \frac{P_1}{P_2} = \left(\frac{V_2}{V_1}\right)^\gamma $$

and you know that $V_2/V_1=0.5$ so you can calculate the new pressure. Then use the ideal gas equation of state:

$$ PV = nRT $$

to calculate the new temperature.

  • $\begingroup$ This is correct only for an adiabatic reversible compression. If the compression is very rapid (irreversible) or non-adiabatic (heat exchanged with surroundings or cup), the result will be different. $\endgroup$ – Chet Miller Aug 16 '17 at 11:38
  • $\begingroup$ Can you explain "where γγ is the ratio of the two specific heats CP/CV." ? Also since pressure would double, and Volume would half, wouldn't the temperature stay the same under the ideal gas law? $\endgroup$ – Robert Short Aug 16 '17 at 21:01
  • $\begingroup$ @RobertShort: see the Wikipedia article on the specific heat ratio for more info on what $\gamma$ is and how it's calculated. Re your second question: all we know for sure is that the volume will halve. You say the pressure will double, but that is incorrect. The pressure will more than double because the temperature will go up. You have to use the equation I cited to calculate the new pressure. $\endgroup$ – John Rennie Aug 17 '17 at 5:01

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