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Calculating work done on gas

  1. Mar 27, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-3-27_14-9-14.png
    upload_2016-3-27_14-9-30.png


    2. Relevant equations
    U = Q + W

    3. The attempt at a solution
    I don't understand why there is no work done on gas from Q to R. I thought that work done on gas = pv?
    If so, then why there is no work done on gas because there is increase in pressure?

    So, I tried to find a reason, and I thought that work done should be the area under the graph, because it is a volume against pressure graph. In this case, then there shouldn't be any work done from P to Q too because the change is a straight line. But there is. There is a work done of +240J.

    Can someone explain to me where I'm wrong? Thanks
     
  2. jcsd
  3. Mar 27, 2016 #2

    ehild

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    Look at the axes. The horizontal one is the pressure, the vertical on is volume. You can imagine that the gas is confined in a cylinder equipped with a movable piston. The gas exerts force on the piston, and does work if the piston moves. If the piston moves, the volume of the gas changes. Does the volume change from Q to R? Is work done without displacement of the piston?
    The work is the area under a P(V) curve. But the curve in the figure is V(P).
    From P to Q, the volume changes from 8*10-4 m3 to 2*10-4 m3 while the pressure stays the same, 4*105 Pa. The work done by the gas is W=PΔV, negative, as the volume decreases. The work done on the gas is 240 J.
     
  4. Mar 27, 2016 #3
    Thanks. But I'm still confused. Why is work done by a gas the area under a P(V) curve? Why must pressure be constant? And how can it be, since ## P ∝ 1/V ## ?
     
  5. Mar 27, 2016 #4

    ehild

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    You should know that the work of the force F(x) during the displacement from x1 to x2 is ##W=\int _{x1}^{x2}{F(x) dx}## The definite integral is defined so as it gives the area of the F(x) curve between x1 and x2,
    The force exerted by the gas on the piston of area A is PA. (P is the pressure). If the piston moves by dx, the elementary work is dW=P(Adx), but Adx is the change of volume dV= Adx, so the elementary work done by the gas is dW=PdV. The entire work when the volume of the gas changes from V1 to V2 is the integral W=##\int _{V1}^{V2}{PdV}##, the area under the P(V) curve between V1 and V2.
     
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