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Homework Help: Thermodynamics help! Ideal gas equation

  1. Jan 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose that you measured the product pV of 1 mol of a dilute gas and found
    that pV = 22.98 L atm at 0°C and 31.18 L atm at 100°C. Assume that the perfect gas law
    is valid, with T = t(°C) + a, and that the value of R is not known.

    Determine R and a from the measurements provided. Express your result for R in J K


    2. Relevant equations

    PV = nRT
    T = t(°C) + a

    3. The attempt at a solution
    Ok I answered the first part:
    R = slope = (31.18 - 22.98 / 100 - 0 )
    = 0.082 L atm C

    what is a ??how do I find it??
  2. jcsd
  3. Jan 17, 2009 #2


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    Staff: Mentor

    What temperature scale should be used in the ideal gas equation?
  4. Jan 17, 2009 #3
    What do you mean?
  5. Jan 17, 2009 #4


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    Staff: Mentor

    What temperature scales do you know?
  6. Jan 17, 2009 #5
    ohh well we usually do it in Kelvin, but the question asks for Celsius, so I guess we dont have to convert anything, no?
  7. Jan 17, 2009 #6


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    Staff: Mentor

    How do you convert between C and K? What is "a"?

    Note, your R value looks correct, but you are just lucky - "a" canceled out.
  8. Jan 17, 2009 #7
    C= value of Temp in C + 273.15 = value in K

    I have no idea what "a" is.

    I don't understand the meaning behind the given equation: T = t(°C) + a

    What's t(°C) ? and how are we supposed to use it to determine "a"?

    I'm so lost. :(
  9. Jan 17, 2009 #8


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    Staff: Mentor

    It is so simpe that you will feel ashamed once you will get it :smile: Compare two equations that you have wrote:

    temp in K = temp in C + 273.15

    T = t(°C) + a

    Note, that T means absolute temperature (Kelvins), while t(°C) means temperature in Celsius degrees. Do you see that it is the same equation?
  10. Jan 17, 2009 #9
    haha ok, so a is 273.15

    so you're basically saying,

    PV = n R (t(°C) + a)

    1) 22.98 = nRa
    2) 31.18 = nR(100 + a)

    If we do 2-1

    31.18 - 22.98 = 100Rn + nRa - nRa
    8.20 = 100R (1 mol)
    R = 0.0820 L atm C

    and if we plug back into one of the equations,
    we get 22.98 L atm= (1mol) (0.0820L atm C)a

    a = 280.24 C ???
    is this our right unit?
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