Thermal Physics

• Han chiang

  7 Jan 11, 21:54

  Hello forumers, I have some questions on Thermal Physics III.

  There is this question where a situation is given and I need to fill in the boxes (delta U, Q, and W) with -, + or 0.

  1) Pan filled with room temperature water over a hot stove. System: Water in pan

      My Q is +, W is - , but I don't know how to find delta U.

  2) Hot water in kettle vaporizing into water vapour as it reached its boiling point. System: Water originally in kettle

      My Q is +, W is - , again I don't know how to find delta U.

   

  Another side question. If something happens rapidly means the Q is 0? e.g. Rapidly pumping up a bicycle tyre - system: air in pump, and air quickly leaking out of a balloon - system: Air originally in balloon.

   

  Thank you.

• wee_ws

  8 Jan 11, 07:48

  Hi,

    The water at room temperature will increase on a hot stove, so internal energy will increase.

    Water undergoes a change of phase from liquid to vapour. For this to happen, there has to be an increase in molecular energy to overcome the attraction that held the molecules together in the liquid state. Thus, internal energy must increase.

    Thanks.

  Cheers,
  Wen Shih

• wee_ws

  8 Jan 11, 07:54

  Hi,

    We may assume that pumping a tyre is an adiabatic change. 

    The laws of thermodynamics associated with the hand pump is not straightforward and much information is lacking in your description. This article may help to shed some light:

    http://www.physics.princeton.edu/~mcdonald/examples/tirepump.pdf

    Thanks.

  Cheers,
  Wen Shih

• Han chiang

  8 Jan 11, 11:45

  Oh, when the water in the pan is on the hot stove, the internal energy is increasing because heat is transferred to the water? And from liquid to vapour state, the water gain potential energy to break the molecular forces. But water at 100 degrees and vapour at 100 degrees has the same KE right, since temperature are the same. This is making use of U = Sum of KE + PE ? Because sometimes use delta U = Q + W . One is minus, one is plus, in the end don't know the sign of delta U. One side question. Let say I am doing a question on ideal gas. I can use delta U = 3/2 delta PV, 3/2 delta nRT or 3/2 delta NkT. Can I also use delta U = Q + W to find the delta U? Because I can only get the answer from the above 3 ideal gas equations, when I use delta U = Q + W I don't get the answer..
• Tama rebirth

  8 Jan 11, 12:22

  @Han chiang

   

  The question depends on what type of gas it is. Yes, it is ideal gas, but is it monatomic or diatomic or what...3/2 is only for monatomic. 5/2 for diatomic and etc etc. 

  Be careful of using delta U = 3/2 delta NkT or the likes....it contains a lot of assumption such as monatomic ideal gas and all. Go back to basic formula first such as First Law of Thermodynamics (delta U = Q + W) . 

   

   

• Kouer78

  9 Jan 11, 10:23

  Hi,

   

  If I am not wrong, delta U = Q + W only apply for ideal gas.

  But assuming can be used in the following case,

  1) Heat supplied water, Q +ve, and common sense tells me temp of water increase, change in U increase, which means W -ve, and work down by system must be less than heat supplied for change in U to be +ve.

   

  2) Change in state, means gain in PE and hence gain in U, for same KE. Gas expand so work done by gas, W -ve. Heat given to gas to change in state, Q +ve.

   

  For adiabatic process, think must have either condition:

  1. thermal insulator (no heat can enter or leave the system)

   

  2. process done very quickly ( reason being process too fast for any significant heat exchange, no time for heat to enter or leave the gas)

   

  Hope it helps. Do correct me if i am wrong.

• wee_ws

  9 Jan 11, 13:09

  Hi,

    Warm welcome to Kouer78 and thanks for your reply :)

    The first law of thermodynamics can be applied to other systems.

    In Giancoli's book, the first law is applied to human metabolism.

    Thanks.

  Cheers,
  Wen Shih

• Han chiang

  9 Jan 11, 13:22

  Ok I get it now. Thank you.

  Another question: There are 3 different paths, abd, acd and ad whereby a system consisting of 4 moles of gas can undergoes from point a to d. Note that path ad is an isothermal curve of temperature T.

  At a, P = 100000Pa, V = 0.5 m^3

  At b, P = 100000Pa, V = 0.2 m^3

  At c, P = 250000Pa, V = 0.5 m^3

  At d, P = 250000Pa, V = 0.2m ^3

  I need to fill up delta U, Q and W. So if I want to find Wabd, I use Wd - Wa correct or not? For delta Uabd, I also use Ud - Ua. Somehow it feels abit weird...

   

• wee_ws

  10 Jan 11, 07:53

  Hi,

    a to b: Isobaric process
    Work done on the gas = P x (delta V) = (1 x 10^5) x 0.3 = 3 x 10^4 J

    b to d: Isovolumetric process
    No work is done, since delta V = 0.

    a to d: Isothermal process
    No change in internal energy, since delta T = 0.

    Hence, by the first law of thermodynamics, we have

    0 = Q + 3 x 10^4 => Q = -3 x 10^4 J.

    Thanks.

  Cheers,
  Wen Shih

• eagle

  10 Jan 11, 17:46

  Wow, I missed a lot of interesting Q&A in this thread :)

• wee_ws

  10 Jan 11, 19:05

  Hi eagle,

    Welcome back :)

  Cheers,
  Wen Shih

• Han chiang

  10 Jan 11, 21:06

  Now I know, U is a state function. So to find delta Uxyz, I can just use final minus initial. But to find Wxyz, I need to look under the PV graph right? Alamak ideal gas and graph abit confusing leh...

  Can you help me check my answer for Wacd also? My Wacd is 75000J.

  Thank you.

• wee_ws

  11 Jan 11, 06:34

  Hi,

    7.5 x 10^4 J is correct! Good effort!

  Cheers,
  Wen Shih