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Why multiply atm by atmospheric pressure in pascals?

  1. Jun 18, 2011 #1
    Consider the following two-step process. Heat is allowed to flowout of an ideal gas at constant volume so that its pressure dropsfrom PA = 2.7 atmto 1.4 atm. Then the gas expands at constant pressure, from avolume of 6.8 L to VC = 13.1 L where the temperature reaches its originalvalue. See Fig. 15-27.

    In the following work, why is 1.4 atm multiplied by 1.013e5 pascals? And to clarify, PA=2.7 just means pressure A.
    Figure 15-27
    (a) Calculate the total work done by the gas in the process (in joules).
    Work done by gas along the parts BC W2= PdV
    = P( Vc-Vb)

    = 1.4*1.013*105( 13.1-6.8)*10-3
    =893.466J
     
  2. jcsd
  3. Jun 18, 2011 #2

    gneill

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    Staff: Mentor

    It's a unit conversion. The equation requires that the pressure be in pascals, and 1 atm = 1.01325 x 105 pascals.
     
  4. Jun 18, 2011 #3
    So to get to joules of work, you have to multiply pascals times volume? And why is the answer divided by a thousand (10^-3)?
     
  5. Jun 18, 2011 #4

    gneill

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    Staff: Mentor

    Because the volume should be in cubic meters, and the values given in liters. How many liters in a cubic meter?
     
  6. Jun 18, 2011 #5
    1 liter = .001 m^3. So pascals times cubic meters equals joules? I'm confused.
     
  7. Jun 18, 2011 #6

    gneill

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    Staff: Mentor

    No need to be confused. Pressure x volume does indeed have the units of energy.
     
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