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Ideal Gas compressed at constant pressure

  1. Mar 11, 2016 #1
    Hi,
    Could I please get some guidance on my approach and solution, for this particular problem.
    Any assistance welcome.

    1. The problem statement, all variables and given/known data

    An ideal gas is compressed at a constant pressure of 1.3 atm from a volume of 20 L to 12 L. During this process it gives off 3.69 kJ of heat. What is the change in internal energy of this gas?

    2. Relevant equations
    W=P.ΔV
    Q = ΔU + W

    3. The attempt at a solution
    Calculate the work done.
    W = P.ΔV = 1.3 atm x (20 -12) L = 10.4 L atm
    Convert to Joules = 10.4 x 101.3 = 1053.5 J
    W = 1053.5 J

    Q = ΔU + W
    Given Q, calculate the change in internal energy.
    ΔU = Q - W = 3.69 x 10^3 - 1053.5 = 2636.5 J

    ΔU = 2636.5 J
     
  2. jcsd
  3. Mar 11, 2016 #2

    TSny

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    Gold Member

    Your approach looks good, but be careful with signs.

    Is ΔV positive or negative? Is W positive or negative? What about the sign for Q?
     
  4. Mar 14, 2016 #3
    Thank you for the feedback. I believe I spotted my mistake.

    The volume was compressed from 20 to 12. So it should be 12-20.
    Which would result in a negative W value which would carry through my calculations.

    Calculate the work done.
    W = P.ΔV = 1.3 atm x (12 - 20) L = -10.4 L atm
    Convert to Joules = -10.4 x 101.3 = -1053.5 J
    W = -1053.5 J

    Q = ΔU + W
    Given Q, calculate the change in internal energy.
    ΔU = Q - W = 3.69 x 10^3 - (-1053.5) = 4743.5 J

    ΔU = 4743.5 J
     
  5. Mar 14, 2016 #4
    "it gives off 3.69 kJ of heat"

    Shouldn't Q be negative?
     
  6. Mar 14, 2016 #5

    TSny

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    As a rough check, do you expect the internal energy to increase or decrease in this process? (What happens to the temperature?)
     
  7. Mar 14, 2016 #6
    Thank you both. I went back and did some more reading!

    In this example the system is giving off heat so, so Q should be negative.
    Also my understanding, is that because the work is done to the system, W should be positive.

    Calculate the work done.
    W = P.ΔV = 1.3 atm x (20 - 12) L = 10.4 L atm
    Convert to Joules = 10.4 x 101.3 = 1053.5 J
    W = 1053.5 J

    Given Q, calculate the change in internal energy.
    ΔU = - Q + W = - 3.69 x 10^3 + 1053.5 = -2636.5 J

    ΔU = -2636.5 J

    So the internal system loses 2636.5 Joules of energy.
     
  8. Mar 14, 2016 #7
    In that case, you must be using the formula Q + W = ΔU.

    Look at it this way. Consider the system to be a gas kept in a cylindrical bowl with a piston on the top. There is a bunsen burner at the bottom. Now, the internal energy can be increased in two ways : you heat the gas up or/and push the piston down. Thus the change in internal energy is given by the above formula. Mind you, this picture is only to help you gauge the formula properly and place the W in the right place. Once you have done so, the corresponding signs of Q, W and ΔU will take care of the situation ( as @TSny so elaborately said in post #2 ).

    By the way, take care in defining W too. Stick to one definition. Do not grapple with "is it work done by the system" or "work done on the system". The only difference this gives you is in the position of W. In my picture above, W is work done on the system and thus appears along with Q. If it was the other way around, then it would appear with ΔU (Can you see why?).

    PS : I tell all this because I struggled with this barely an year ago.
     
  9. Mar 14, 2016 #8
    If you are using the equations Q = ΔU + W and dW=PdV, then these equations imply that work done by the system on the surroundings is positive, and work done by the surroundings on the system is negative.
     
  10. Mar 15, 2016 #9
    Thank you all for the responses but I got a bit confused.
    However it's clear that this confusion was just waiting in the background to emerge... so thanks, this has been very helpful.

    Hopefully, the following is correct.

    ΔU = Q + W (I'm go to use this formula as I find it more intuitive).
    Q = is negative in this example as heat is given off by the system
    W = work done to the system, so W is positive

    Calculate the work done.
    Change in volume is 8L.
    W = P.ΔV = 1.3 atm x 8 L = 10.4 L atm
    Convert to Joules = 10.4 x 101.3 = 1053.5 J
    W = 1053.5 J

    Calculate the change in internal energy.
    ΔU = - Q + W = - 3.69 x 10^3 + (1053.5) = -2636.5 J
    ΔU = -2636.5 J

    So the heat loss exceeds the work gain, so the change in internal energy is negative.
     
  11. Mar 15, 2016 #10
    You really need to be careful about how you handle signs. You just can't go on your intuition. If you are using the equation Q+W=ΔU, then dW=-PdV. And the sign of Q can't suddenly change from positive to negative. In this problem, Q = -3.69 x 10^3. You got the final answer right, but your chances of getting the answers to future problems correct are very low if you continue with your slipshod handling of the mathematics. Please try to be much more exact in how you handle the signs of terms.
     
  12. Mar 15, 2016 #11
    So you are saying that W should be negative.
    Therefore the equation should become: -Q+(-W) = ΔU (as Q is negative due to heat loss)

    That would mean the answer to this question would be :
    ΔU = - Q + (-W) = - 3.69 x 10^3 + (-1053.5) = -4743.5 J
    ΔU = -4743.5 J

    Which isn't the correct answer.
     
  13. Mar 15, 2016 #12
    No. I said that you had the right answer, but I can see that you are still confused.

    If you are using the equation Q + W = ΔU, then, in this problem W = -PΔV= - 1.3 atm x (12-20) liters = - 1.3 x (-8) = + 10.4 liter atm = + 1053.5 J

    Since the gas gives off heat to the surroundings, Q = - 3.69 x 10^3 J.

    So Q + W = -3690 + 1053.5 = -2636.5 J = ΔU

    This is the same answer you got in post #9, but it was obtained by handling the signs in a consistent and mathematically correct way.
     
  14. Mar 15, 2016 #13
    Understood, thanks for the response.
     
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