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Homework Help: HI I'm new here. I've been having a little bit of trouble with a few

  1. Nov 15, 2012 #1
    HI I'm new here. I've been having a little bit of trouble with a few of my physics problems. It seems like very basic stuff. I went to my professor's office hours and he didn't help me. He told me the wrong stuff, and when I emailed him back, he said different stuff.

    An ideal monatomic gas expands isothermally from 0.550 m^3 to 1.25 m^3 at a constant temperature of 610 K. If the initial pressure is 1.10*10^5 Pa.
    (a) Find the work done on the gas.

    (b) Find the thermal energy transfer Q.

    The internal energy is 0J, therefor -W=Q. I tried nRT*ln(V_f/V_i) but that didn't come out. I'm not really sure what else to do.
    2) A gas is enclosed in a container fitted with a piston of cross-sectional area 0.150 m2. The pressure of the gas is maintained at 5800 Pa as the piston moves inward 21.5 cm.

    (b) If the internal energy of the gas decreases by 8.60 J, find the amount of heat removed from the system by heat during the compression.

    I calculated the work done by using W=F*A*d, which comes out to be around -187. From there I tried using delta U = W + Q. Which would be -8.60=.187 + Q. However, my Q isn't correct, it is however within 10% of the correct answer.
  2. jcsd
  3. Nov 15, 2012 #2
    Re: Thermodynamics

    First of all I use the convention:
    Q = U + W
    If the gass expands, W is positive

    1) W = nRT ln(Vf/Vi) = PV ln(Vf/Vi) = 1.10*0.550*105 ln(1.25/0.55) = 49.6 kJ
    Q = W = 49.6 kJ

    2) You are using your convention wrong. if U = W + Q, W is negative when gas expands, so, in this conventiuon, W = +187J and Q = -8.6 - 187 = -195.6J

    You can use Q = U+W and it would give the same result
    But in this case W = -187J, Q = -195.6J
  4. Nov 15, 2012 #3
    Re: Thermodynamics

    Thanks for the help on those two. Any idea how to solve the work from these two graphs?
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