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Pressure of gas and work done

  1. Jul 8, 2011 #1
    PV = nRT

    When a gas in a cylinder expands, the gas pushes the piston upward against a constant opposing external atmospheric pressure P. The work done by the gas on the surroundings is w= -P[itex]\Delta[/itex]V

    [itex]\Delta[/itex]V is the change in volume

    My question is:
    Assuming the Temperature stays constant (as my book does), then as the gas expands (the volume increases) then the Pressure must also decrease, right? bc PV = nRT and everything else is constant....

    But pressure is a force, and according to Newton, every force has an equal and opposing force. So the pressure that the gas exerts must equal the atmospheric pressure, right?
    But how can this be true if the pressure of the gas is decreasing as the volume is increasing?
     
  2. jcsd
  3. Jul 8, 2011 #2
    Let's suppose T and n fixed. If you don't apply any other forces from the outside besides the atmospheric pressure, the expansion can happen only if the pressure of the perfect gas is greater than the atmospheric pressure.
    The work done by the gas can be correctly calculated using your formula, but in this case p is not the pressure of the gas during the expansion. You're right when you say that in this case if the volume increases the pressure decreases.
    If the pressure of the gas is equal to the atmospheric pressure, the system is in equilibrium and there would be no expansion.
     
    Last edited: Jul 8, 2011
  4. Jul 8, 2011 #3

    I'm confused because of Newton's law where it says that the force of A on B is equal and opposite to the force of B on A.
    So shouldn't the pressure of the gas on the atmosphere equal the pressure of the atmosphere on the gas?
     
  5. Jul 8, 2011 #4
    Let's think the cylinder as if it has one base free to move. The gas inside the cylinder applies a pressure p on the base (so the force is F_in=p S, where S is the area of the base). Because of Newton's law, the base itself applies on the gas the same force (in opposite direction). Moreover, the gas outside the cylinder applies a pressure p_atm on the base (so the force this time is F_out= - p_atm S; the minus sign is needed because the two forces are opposite in direction), while the base applies on the outside gas (- F_out). Newton's law is not violated. In conclusion, on the base the total force is F=F_in - F_out=(p - p_atm) S, so the expansion happens when the two pressures are different.

    In other words, the two forces (equal and opposite) considered by Newton's law are not to be applied on the same object (otherwise they always cancel each other and there would be no motion at all).

    ps Sorry for my English, I hope I made myself clear enough.
     
  6. Jul 8, 2011 #5
    Remember these forces act on a piston. The piston may have a mass 'm' or be considered mass-less. The force applied by the gas on the piston is P*A, but since P is a function of volume like you stated, the force is a variable force. The other force applied on the piston is from the atmospheric pressure and is Patm*A. Notice in reality there would also be a friction force between the piston and the cylinder wall. So summing forces on the piston would result in:

    A*(nRT/V)-Patm*A = m*a

    Now if m=0 or a=0 then you arrive at the steady state where the pressure of the gas is equal to atmospheric pressure.
     
  7. Jul 8, 2011 #6
    Wow. Yes it is clear now. Thank you!
     
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