1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Specific Heat Capacities in co-ordination with Thermodynamics

  1. Jul 8, 2014 #1
    Please help me qualitatively in the following points :
    1) If in a system(consider a cylinder) fixed with a piston , if the piston is moved suddenly then how can a process be adiabatic.
    2) I understood that the process would be irreversible but, if the process is adiabatic then is the relation PV^y = constant (where y is gamma, P is pressure, V is volume) is true for irreversible process too. (In many books they have written that the relation is true only for reversible processes)

    In the following question , I am not getting the real essence of the mechanics of the process. Please explain:

    A gas is enclosed in a cylindrical can fitted with a piston. The walls of the can are adiabatic. The initial pressure, volume and temperature of the gas are 100 kPa, 400 cc (cubic cm) and 300 K respectively. The ratio of the specific heat capacities of the gas is Cp/Cv=1.5 . Find the pressure and temperature of the gas if it is (a)suddenly compressed to 100 cc (cubic cm). (b)slowly compressed to 100 cc (cubic cm).

    Here, the answer to both the cases is given same by taking PV^y = constant (where y is gamma). How would it be??
  2. jcsd
  3. Jul 8, 2014 #2

    Philip Wood

    User Avatar
    Gold Member

    First question. There's very rapid compression and there's rapid compression…

    If you move the piston in at a speed that is not very much smaller than the mean (or the rms) speed of the molecules, then you are compressing the gas irreversibly. There is a local large rise in pressure near the piston, and that rise will never be reproduced when you move the piston back out.

    Now go to the other extreme. Move the piston out at, say, 0.001 m/s. As the gas does work on the piston it will lose internal energy, lowering its temperature. But as soon as this happens, heat will flow in from the surroundings (assuming that their temperature is equal to the initial temperature of the gas), and the temperature will fall hardly at all before the heat coming in through the walls of the cylinder balances the work being done by the gas. So, to all intents and purposes the gas temperature stays constant (isothermal expansion) as the gas does work - provided the work is done slowly.

    Between these two extremes of speed of movement of the piston there lies a range of speeds which are too low for significant irreversibility, but too high to give enough time for significant heat to flow in from the surroundings. We have (nearly) irreversible adiabatic expansion
  4. Jul 9, 2014 #3

    Philip Wood

    User Avatar
    Gold Member

    Sorry: "reversible" not "irreversible" in last line of previous post.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook