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Homework Help: Conceptual problems-Work done,KE and internal energy

  1. Jul 3, 2013 #1
    1. The problem statement, all variables and given/known data

    Q.1)A gas cylinder is lifted from the ground floor to the first floor.What is the amount of work done on the gas?What is the amount of work done by the gas?Is the internal energy of the gas increased?

    Q.2)When we place a gas cylinder on a car and the car moves,does the kinetic energy of the molecules increase?Does the temperature increase.?

    2. Relevant equations

    3. The attempt at a solution

    Reasoning 1) The work done on the gas or by the gas is given by pΔv,which is zero in both the cases.The internal energy of the gas shouldn't change.But then the work done by the external agent changes the potential energy of the CM as the cylinder moves upwards.How do we see this in light of the work done by the external agent?

    Reasoning 2)Since neither the heat is supplied ,nor there is change in the volume of the cylinder,the internal energy of the gas should remain same.So,the kinetic energy of the molecules should also remain the same.

    Is my reasoning correct?
  2. jcsd
  3. Jul 3, 2013 #2


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    NO. The work done on the gas is the external force multiplied by the displacement it causes. The system raised to the next floor, the height changed with h. There are two external forces: gravity and the applied force. All do work.
    The volume of the gas does not change, the gas does not do any work.

    Work of the applied force is Fh, the work of gravity is -mgh. The work of all forces is equal to the change of KE. If the forces are of equal magnitude the KE does not change. Fh-mgh=0. Gravity is a conservative force, its work is equal to the negative change of the PE. PE=mgh, so the work of the applied force Fh=mgh=ΔPE.

    The KE of the random motion of the molecules remains the same, and also their total KE remains the same, if the applied force just balanced gravity.

    So the sum of the KE of all molecules changes if the system gains some velocity as a whole, but this macroscopic velocity does not change the KE of random motion.

  4. Jul 4, 2013 #3
    Referring to Q1)

    -Since the change in KE is zero ,the net work done on the gas is zero .Of course,the work done by applied force and gravity are non-zero.Am I correct?

    -How do we calculate work done by gas ? Is it given by pΔv ?Since Δv=0,the work done by gas is zero.Right?

    -Since net work done and heat supplied is zero,the internal energy remains same.Is that correct?
  5. Jul 4, 2013 #4


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    When you speak about work you have to specify by what and on what the work is done.
    Work by the gas on what?
    Work by expanding gas on the surroundings is pdV only for a quasi-static process, when the gas is in equilibrium with its surroundings. Pressure would not be defined otherwise. If you let the gas expand in vacuum, it does no work.

    The whole system - the cylinder filled with gas - exerts force on the external agent and also on the Earth. If you push the cylinder with force F it pushes your hand with the same force. As your hand moves, work is done on it by the cylinder of gas.
    In the same way, the cylinder of gas exerts force and does work on the Earth.

    The cylinder of gas has got some potential energy as a whole. But that does not change the internal energy - the energy of random motion of the gas particles. If the positions of all molecules change by the same, the CM shifts, but the distribution of the particles in the vessel stays random with respect to the CM.

  6. Jul 4, 2013 #5
    Do we distinguish between work done by gas and work done by cylinder filled with gas .

    If yes,the work done by the two would differ.Is that so?
  7. Jul 4, 2013 #6


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    Ignoring gravitational interaction between the gas molecules and the Earth, the gas molecules in the cylinder interact only with the walls. So they can do work on the cylinder only. If the cylinder has rigid walls, no work is done by the gas.

    The cylinder and the gas inside make a system - an object. It can do work on other objects it interacts with.

  8. Jul 4, 2013 #7
    Well...I am glad to put my doubts in front of you...You have beautifully removed my misconceptions.

    Thanks a lot for your time and patience.Your work done on me has energized me quite well :tongue:
  9. Jul 4, 2013 #8
    What about Q2? The kinetic energy of the molecules is only defined for the cylinder frame. If the car moves with a constant velocity the then there will be no effect of its motion on the molecules. But if the car accelerates then a pseudo force will act on each gas molecule and will increase the velocity of each, so KE will increase. But a steady state with the walls balancing the pseudo force will result eventually (I think) and the KE will become constant.
  10. Jul 4, 2013 #9


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    You can define KE of a molecule in the ground frame, and you can define their sum, as the KE of the whole system, a cylinder of gas.
    When the car goes with constant velocity V, the cylinder moves together with the car, and CM of the gas molecules moves together with the cylinder. The gas molecules have their random velocity with respect to the CM, it is vri of the i-th molecule, so their velocity with respect to the ground is V+vri. The sum of the KE of the molecules is Ʃ(0.5(V+vri)2)=0.5V2M+Ʃ(0.5mvri2), where M is the total mass. See https://www.physicsforums.com/showthread.php?t=699878 post#11

    You can say that the molecules feel the pseudo-force in the accelerating cylinder, or the cylinder accelerates with respect to the CM of the system of molecules. The molecules colliding with the the front wall get less backward impulse, those colliding with the back wall, get greater forward impulse, so the magnitude of their momenta changes. The wall exerts some forward force on the ensemble of molecules. At the end the average velocity becomes the same as that of the container, or we can say that the CM of the system of particles moves together with the cylinder.
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