1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Isobaric Expansion

  1. Sep 27, 2007 #1
    1. The problem statement, all variables and given/known data
    5.00 g of nitrogen gas at 22.0 deg C and an initial pressure of 2.00 atm undergo an isobaric expansion until the volume has tripled.

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

    2. Relevant equations

    Equation of Isobaric thermodynamic: W = -p(delta V)
    Ideal Gas Law: PV=nRT

    3. The attempt at a solution
    Get the number of moles of the N2 gas: N2 = 28 g/mol ; moles = 5/28
    Convert 22 deg C to Kelvin: 273.15 + 22 = 295.15 K
    Convert 2.00 atm to Pascals: 2 atm = 202 650 pascals
    R is 8.31
    Find V initial:
    PV = nRT
    (202650)V = 0.1786(8.31)295.15
    Vi = 0.0022
    Find V final: Vf = 3(Vi) = 0.0065

    Now Find the Work done:
    W = -p(DELTA V)
    Since pressure stays constant at 202,650 Pascals then:
    W = -202650(0.0065-0.0022) = 871.395 J

    However this is wrong, and I do not understand why. Can someone walk me through this problem?
  2. jcsd
  3. Sep 27, 2007 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Start with the first law. Since P is constant:

    [tex]\Delta Q = \Delta U + P\Delta V[/tex]

    Since [itex]\Delta U = nC_v\Delta T[/itex]

    [tex]\Delta Q = nC_v\Delta T + P\Delta V[/tex]

    Find the change in T using PV = nRT to find the heat flow [itex]\Delta Q[/itex]

    Last edited: Sep 27, 2007
  4. Sep 27, 2007 #3
    I'm not familiar with the procedure for this type of question, but it says that 2.00 atm is the initial pressure. I would guess that it would change.

    However, as I said, I don't know. Sorry if this is wrong.
  5. Sep 27, 2007 #4
    Thanks for directing me to the First Law of Thermodynamics. I have never thought of using the relationship.

    However, for the record, I did realize that it should be C_p not C_v as it is the specific heat of the gas at constant pressure.
  6. Sep 27, 2007 #5

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    You can certainly use Cp to find the heat flow (Cp takes into account the work done in the expansion):

    [tex]\Delta Q = nC_p\Delta T[/tex]

    Change in interal energy is ALWAYS [itex]\Delta U = nC_v\Delta T[/itex].

    If P is constant:

    [tex]\Delta Q = nC_p\Delta T = nC_v\Delta T + P\Delta V[/tex]

    But [itex]\Delta V = nR\Delta T/P[/itex] so

    [tex]\Delta Q = (nC_v + nR)\Delta T[/tex] so

    [tex]C_p = C_v + R[/tex]

  7. Sep 27, 2007 #6


    User Avatar

    Staff: Mentor

    The problem states "isobaric expansion", so pressure does not change. The volume does change as also stated.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Isobaric Expansion
  1. Isobaric expansion (Replies: 1)

  2. Isobaric expansion (Replies: 5)