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Work done in an Isobaric expansion

  1. Apr 21, 2007 #1
    1. The problem statement, all variables and given/known data

    4.20g of nitrogen gas at 22.0 C and an initial pressure of 2.30 atm undergo an isobaric expansion until the volume has tripled.

    A.)How much heat energy is transferred to the gas to cause this expansion?

    b.)The gas pressure is then decreased at constant volume until the original temperature is reached. What is the gas pressure after the decrease?

    c.)What amount of heat energy is transferred from the gas as its pressure decreases?

    2. Relevant equations


    W=-p(delta V)

    3. The attempt at a solution

    Ok so here is all of the steps that I take. I know it seems like I do more then I need to but I get all of the information I can in these problems so that I can memorize the ways to do it so here goes.

    First the infromation that I have is

    M = .0042kg
    Ti = 295K (22+273)
    Pi= 2.3 atm = (101325 * 2.3) = 233047.5
    An= 14u = (14 * 1.661 x 10^-27) = 2.33 x 10^-26 Kg
    M/kg = 2.33 x 10^-26 * 6.02 x 10^23 = .014 Kg
    n= .3 mol
    Vi= (.3 * 8.31 * 295)/233047.5 = .00315 m^3
    Vf= .00315 * 3 = .00947

    This is where I'm having a hard time. The equation I have tried using is

    W=-p(delta V)

    -233047.5 * (.00315 - .00947) = 1472.86 (this answer is wrong)

    So what am I doing wrong?

    So here is some more work that I have doen.

    W=nR(deltaT) = 590.79
    Tf= 590.79 + 22 = 612 C = 612 +273 = 885 K (That seems a little hot I'm pretty sure its not right but I can' get any other answer.)
    Last edited: Apr 22, 2007
  2. jcsd
  3. Apr 21, 2007 #2


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    Homework Helper

    There is a MUCH easier way to convert mass into moles!
    #moles = mass of substance / atomic weight of molecule
    Your n value is wrong. Remember that there are TWO nitrogen atoms in a molecule of nitrogen gas.
    Last edited: Apr 21, 2007
  4. Apr 21, 2007 #3
    ok then that would make my n

    n= .15 mol
    Vi= .00158 m^3
    Vf= .00474 m^3
    W= 736.43 (which is also a wrong answer.)
    I used W=-p(Vi-Vf)
  5. Apr 21, 2007 #4
    I am sorry I can't really follow the work. Please include the units.

    I would approach this problem this way...

    First of all, I would sketch a P-V diagram. It starts at some pressure and volume and then undergoes an isobaric expansion tripling its side. With p on your vertical axis and v on horizontal axis on the diagram, draw the a horizontal line a across. If PV=nRT and p is constant P(3V)=nR(3T) right?
    Knowing this, we know that if volume (3v) is constant the temperature drops back to T, P must drop to 1/3 P. Draw a vertical line down. The temperature is the same as the beginning so it will under an isothermal compression back to the original point.

    Are you familiar with the first law of thermodynamic. du=dQ-dw where du=nC_p*dt and dw=n*C_v*dt? You are right that dw = p*dv


    the nitrogen undergoes an isobaric expansion to three times its original volume. nC_p*dt = n*C_v*dt - p*dv. then dq=n*C_v*dt = nC_p*dt+p*dv

    b) We know that the compression is isochoric, meaning volume is not changed. Work is area under the curve so it is zero. du=dQ = n*C_v*dt

    c) This is the isothermal compression. Again, use du= dQ-dw.
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