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Heat physics help

  1. May 7, 2006 #1
    Sir,
    A volume V and pressure P diagram was obtained from state 1 to state 2 when a given mass of a gas is subjected to temperature changes. Based on this graph it is interpreted that, the gas is heated in the beginning and cooled towards the end. Can you please explain this?
     
  2. jcsd
  3. May 7, 2006 #2
    I am herewith attaching the P-V diagram.
     

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  4. May 7, 2006 #3

    Andrew Mason

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    How is PV related to temperature? What is PV in terms of area on the graph? How does that area change as you move along the graph? What does that tell you about how T changes?

    AM
     
  5. May 10, 2006 #4
    Sir,
    I didnt understand.As we move down the graph the area decreases.
     
  6. May 10, 2006 #5

    Andrew Mason

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    What does the rectrangle defined by height P and width V represent? (hint: ideal gas law). If that area decreases from 1 to 2, what does that tell you about temperature at 2 compared to temperature at 1?

    AM
     
  7. May 10, 2006 #6
    According to gas law, PV proportional to T. So greater PV value when in state 1 than in state 2. So T in state 1 is greater than that in state 2. So heating takes place in beginning and cooling in the end. Is it right?
     
  8. May 10, 2006 #7

    Andrew Mason

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    You can tell by eye that the temperature at 1 is greater than at 2 (by comparing PV at those points). In order to see what happens in between, you have to measure how PV (the area of the rectangle whose top right corner touches the line) varies as you move down the line.

    Plot PV=nRT1 or P = nRT1/V. That gives you the P-V curve for constant temperature T1 (isotherm). Where the line is above the isotherm, temperature is higher than T1. Where the line is lower than the isotherm, temperature is lower than T1.

    AM
     
  9. May 11, 2006 #8
    Sir,
    The P-V diagram here which is a straight line doesn't look like the P-V diagram of an ideal gas as for an ideal gas it a curve. So can we apply the gas law in this case?
     
  10. May 11, 2006 #9

    Andrew Mason

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    Yes. The PV diagram for an ideal gas can be anything - it depends on what is happening to it. If temperature is constant it is a curve P=K/V. If it is adiabatic it is another curve: [itex]P = K/V^\gamma[/itex]. If something else is happening to it, it will look like something else (eg a straight line P = KV).

    The purpose of this question is to find out what is happening to the gas from its PV diagram!

    AM
     
  11. May 11, 2006 #10

    Gokul43201

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    The visual inspection method (looking at areas of rectangles) suggested by AM will give you a feel for what you are looking for (ie : the area of the rectangle first increases and then decreases).

    The other way is to solve mathematically, and be sure of this.

    PV = kT for an ideal gas. You want to find what happens to T as V increases from V1 to V2, ie : you want to find dT/dV along the path shown in the figure.

    The given path can be described by P = P' - mV (P', m are positive constants).

    Substitute this into the ideal gas law and find where dT/dV > 0 and where dT/dV < 0.
     
  12. May 12, 2006 #11
    Sir,
    I didn't understand the point you made that the area of the rectangle first increases and the decreases. Could you please explain it in detail? I thought it decreases when we move from state 1 to state 2. I have constructed 2 rectangles in which it seems that the area is decreasing. Please help me in understanding the concept.
     

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  13. May 12, 2006 #12

    Andrew Mason

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    The rectangle PV is the rectangle with corners at (0,0), (0,P), (PV), (V,0). That area is proportional to T. As you move down the line, the area first increases then it decreases. Where it increases, T increases. Where it decreases T decreases. Where the area (0,0), (0,P), (PV), (V,0) > area of (0,0), (0,P1), (P1V1), (V1,0), T > T1. Where the area (0,0), (0,P), (PV), (V,0) < area of (0,0), (0,P1), (P1V1), (V1,0), T < T1.

    Or you can plot the isotherm P = nRT1/V. For the portion of the line which is above the curve P = nRT1/V, T > T1. For the portion of the line that is below the curve P = nRT1/V, T < T1.

    AM
     
    Last edited: May 12, 2006
  14. May 12, 2006 #13

    Gokul43201

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    Compare the sizes of the 3 rectangles. Also look at a general isotherm given by the red line (a hyperbola, xy=const).

    In any case I still recommend you do the math and find out what happens.
     

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