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First and Second Law of Thermodynamics

  1. Feb 1, 2005 #1
    An ideal gas is confined within a thermally isolated cylinder. It consists of N atoms initially at a pressure of p_0. A movable piston seals the right end of the cylinder. A given amount of heat Q is slowly added to the gas, while the piston allows the gas to expand in such a way that the gas's temperature remains constant at T_0.

    Is the internal energy of the gas the same before and after Q is added?
    Is this true since it has to follow the law of the conservation of energy?

    Does the second law of thermodynamics forbid converting all of the absorbed heat Q into work done by the piston?
    Is the movable piston a cycle? I think it is, so it must follow the second law?

    Is the total work done by the gas independent of the area of the piston?

    Well, the total work is dW = p*dV, and since V = A*h, then work is dependent and not independent?
  2. jcsd
  3. Feb 1, 2005 #2


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    What does James Prescott Joule's experiment tell u about ideal gases...??

    What??What kind of the process is described in the problem...?

    Think again...What is changing in the volume...??The area or the "length" of the tube (improperly said)?

  4. Feb 1, 2005 #3

    Andrew Mason

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    Use the Ideal gas law: PV=nRT = internal energy of the gas. If T and n do not change then PV cannot change. (Since V increases, P must decrease as 1/V).

    This is an interesting question. Since the container is thermally isolated, no heat can escape. The first law says that [itex]\Delta Q = \Delta U + \Delta W[/itex]. Since [itex]\Delta U = \Delta PV = nR\Delta T = 0[/itex] in this process, it would appear that [itex]\Delta Q = \Delta W[/itex] where [itex]\Delta W[/itex] is the work done by the system. Since this would violate the second law, we can conclude that this is not the entire system. I think the second law tells us that there had to be an external source of energy, in addition to the added heat, to move the piston and expand the gas. The heat alone could not do the work of moving the piston.

    I don't think so. The total work done by the gas is:

    [tex]\int_{V_i}^{V_f} PdV = \int_{V_i}^{V_f} \frac{dV}{V} = ln(\frac{V_i}{V_f})[/tex]

    It is independent of the area of the piston.

  5. Feb 10, 2005 #4
    While it is true that V = A*h, it is also true that p = F/A. So the A's cancel each other out and the work done is therefore independent of the area. :smile:

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