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Ideal Gas law lab

  1. Nov 23, 2006 #1
    For the first part of my lab, I have a graph with pressure(atm) as a function of temperature(K).

    For the second part of my lab, I have a graph with pressure(atm) as a function of inverse volume(mL^-1)

    For both of these, i am supposed to find the value for R, the gas constant.

    So far far part a i have: PV = nRT
    P = (nR/V)T (in the form y = mx +b),
    So the slope of my graph is nR/V
    I also know that n is kept constant, because it was a closed system and 1mol= 22.4L. And that the initial volume is 20mL.
    And i think I should I pick two points on the line to find the slope and set that equal to nR/V.

    And this is where i'm having trouble, i get some value for the slope, m, so
    m = nR/V
    R = mV/n, here i'm confused at how to find V and n in order to solve for R.

    For part b; PV = nRT
    P = (nRT) 1/V
    So the slope of my graph is nRT
    I did the same thing as part one;
    m = nRT
    R = m/nT, since T was not chnged, I can do that part, but once again, I am confused by finding n.
  2. jcsd
  3. Nov 23, 2006 #2

    Andrew Mason

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    As I understand your lab experiment you are doing the following:

    1) You keep Volume constant, and take measurements of Pressure at various Temperatures. A plot of P as a function of T should be a straight line P = nRT/V with slope m = nR/V and y intercept b=0.

    2) Keeping T constant, take measurements of Volume at various Pressures. A plot of P as a function of 1/V should be a straight line: P = nRT(1/V) with slope m = nRT and y intercept b=0.

    In 1) you determine V by measurement. You just have to keep n and V constant while you change the temperature.

    In 2) you just have to keep n and T constant as you change the volume.

    To find n, you have to weigh the gas. If you use the fact that 1 mole occupies 22.414 L at STP you are indirectly using the known value for R:

    R = PV/nT = 101325*.022414/1*273.15 = 8.3145 J/mol K

    Last edited: Nov 23, 2006
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