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Heat energy of a monatomic gas

  1. Feb 2, 2009 #1

    TFM

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

    One mole of a monatomic ideal gas, initially at temperature T, undergoes a process in which its temperature is quadrupled and its volume is halved. Draw a PV diagram of this process. Then find the amount of heat Q transferred to the gas. It is known that in this process the pressure was never less than the initial pressure, and the work done on the gas was the minimum possible.

    2. Relevant equations

    Ideal Gas Law: pV = nRT

    Work = pdV

    Heat Capacity:

    tex] C = limit \Delta T --> 0 \frac{Q}{\Delta T}[/tex]


    3. The attempt at a solution

    This seems a very simple question, so I feel silly for needing to ask about it, but, I cannot seem to remember an equation for Q. :frown:

    I have stated above some equations I feel may be relevant.

    Any helpful ideas???

    Thanks in Advance,

    TFM
     
  2. jcsd
  3. Feb 2, 2009 #2

    Andrew Mason

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    From the first law:

    [tex]\Delta Q = \Delta U + W = \int nC_vdT + \int PdV[/itex]

    You can determine the change in internal energy from the change in temperature. You can determine the work from the area under the PV graph but first you have to determine the graph. (Be careful because the work is work done ON the gas, so it is negative).

    In order to determine the graph, you have to make some assumptions. The first assumption would be that heat is not removed from the gas in any part of the process. (Otherwise, theoretically you could create a process in which there was no work done on the gas at all - just reduce the temperature to 0, compress and add heat). What would the path of least work involve?

    AM
     
  4. Feb 2, 2009 #3

    TFM

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    Okay, so for the Internal Energy:

    [tex] \int n c_v dT = \int^{4T}_T 1 \frac{3}{2}nR dT [/tex]

    [tex] \int n c_v dT = \int^{4T}_T \frac{3}{2} * 8.31 dT [/tex]

    [tex] \int n c_v dT = \int^{4T}_T 12.465 dT [/tex]

    [tex] \int n c_v dT = [12.465T]^{4T}_T [/tex]

    [tex] \int n c_v dT = [12.465(4T) - 12.465(T)] [/tex]

    [tex] \int n c_v dT = [49.86T - 12.465T] [/tex]

    [tex] \int n c_v dT = 37.395 J [/tex]

    Would the path of least work just be quadrupling the temperature, then halving the volume?
     
  5. Feb 2, 2009 #4

    Andrew Mason

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    I think you mean 37.395 T. (round to 3 sig. figs or 37.4 T).

    No. Besides, if you raise the temperature to 4T and then do the compression it will get even hotter.

    Plot the PV diagram for the adiabatic path (no heat lost or gained) to V = 1/2V0. How does the work (area under the path) compare to paths above and below the adiabatic? If we eliminate the ones below the adiabatic (since we are assuming that no heat flows out of the gas) what does the adiabatic represent in terms of the amount of work relative to the higher paths?

    AM
     
  6. Feb 2, 2009 #5

    TFM

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    Well an adiabatic process is where no heat enters or leaves the system. The Adiabatic also represents that the work done is equal to the increase in internal energy.

    You also have the equations:

    [tex] TV^{\gamma - 1} = c_1 [/tex]

    [tex] PV^{\gamma} = c_2 [/tex]

    [tex] \frac{T^{\gamma}}{P^{\gamma - 1}} [/tex]

    Is this useful?
     
  7. Feb 3, 2009 #6

    Andrew Mason

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    Yes. The integration for the work done in an adiabatic process is set out here: (see here). It seems to me that, if no heat flows out of the system, the path that uses the least amount of work would be the adiabatic path to V/2 and then add heat at constant volume.

    To find the work using [itex]\Delta U = -W[/itex] (Q = 0) use:[tex] TV^{\gamma - 1} = c_1 [/tex]

    You can determine the temperature at the end of the adiabatic compression and, therefore, the change in internal energy (= work done). You then just have to work out Q in raising the temperature from that up to 4T

    AM
     
    Last edited: Feb 3, 2009
  8. Feb 3, 2009 #7

    TFM

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    OKay, so I need to firstly use:

    [tex] W = int^V_{1/2V}p dv [/tex]

    rearranging:

    [tex] TV^{\gamma - 1} = K [/tex]

    to give:

    [tex] v = (\frac{K}{T})^{1/(\gamma - 1)} [/tex]

    and then insert this into

    [tex] W = int^V_{1/2V}p dv [/tex]

    ???

    Then I have to use

    [tex] C = \frac{Q}{\Delta T}[/tex]

    to work out Q for the increasing Temperature?

    Does this seem okay?
     
  9. Feb 4, 2009 #8

    TFM

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    Okay, i think the previous post was wrong, i misunderstood the link's importance. What I should have done was:

    [tex] TV^{\gamma - 1} = C [/tex]

    gamma (monatomic Ideal gas = 1.66)

    [tex] TV^{1.66 - 1} = C [/tex]

    [tex] TV^{0.66} = C [/tex]

    [tex] T_1 V^{0.66} = T_2 (\frac{V}{2}^{0.66}) [/tex]

    Take logs:

    [tex] ln(T_1) + 0.66ln (V) = ln(T_2) + 0.66ln (V) - 0.66 ln(2) [/tex]

    Rearrange:

    [tex] ln(T_1) - ln(T_2) = 0.66ln (V) - 0.66ln (V) - 0.66 ln(2) [/tex]

    [tex] ln(T_1) - ln(T_2) = -0.66 ln(2) [/tex]

    [tex] ln(T_1 / T_2) = -0.66 ln(2) [/tex]

    take exponentials:

    [tex] \frac{T_1 }{T_2} = (\frac{1}{2})^{0.66} [/tex]

    Does this look better?
     
  10. Feb 5, 2009 #9

    Andrew Mason

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    Yes, apart from the unnecessary algebra that you are using:

    [tex] T_1 V^{0.667} = T_2 \left(\frac{V}{2}\right)^{0.667} [/tex]

    [tex]\frac{T_2}{T_1} = \left(\frac{V}{V/2}\right)^{0.667} = 2^{.667} = 1.59[/tex]

    That gives you the temperature (T2) at the end of the adiabatic process (1.59T) and, therefore, the work: [itex]W = -\Delta U = -nC_v(.59)T[/itex]. You then raise the temperature at constant volume until the temperature reaches 4T, and you can work out the heat flow:

    [tex]\Delta Q = C_v(4-1.59)T[/tex]

    AM
     
  11. Feb 5, 2009 #10

    TFM

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    OKay, so that makes sense, but where does the 37.4 T fit we found earlier?
     
  12. Feb 5, 2009 #11

    Andrew Mason

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    That is the total change in internal energy (the energy is 37.4T, units are joules and T in Kelvins) . It comes from the two different processes: 1. work done in compressing the gas to V/2 and 2. heat flow in raising temperature at constant volume. So it does not tell you how much is from heat and how much is from work done on the gas.

    AM
     
  13. Feb 5, 2009 #12

    TFM

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    Okay, So I presume I need to use:

    [tex] \Delta Q = U_{total} + W_{total} [/tex]

    And:

    [tex] \Delta Q = 37.4T + W_{total} [/tex]

    [tex] \Delta Q = C_v(4-1.59)T [/tex]

    [tex] C_v(4-1.59)T = 37.4T + W_{total} [/tex]

    And

    [tex] C_v(4-1.59)T = 37.4T + (-nC_v(.59)T) [/tex]

    Does this look right?
     
  14. Feb 7, 2009 #13

    Andrew Mason

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    Cv is the molar specific heat. To find the total heat you have to multiply by n. In this case n=1 so it does not change the answer.

    Just a problem with the n in the formula for [itex]\Delta Q[/itex].

    You can check this yourself to see that the Q and W add up to the change in internal energy (37.4T Joules).

    [tex]W = -\Delta U_{adiabatic} = -nC_v\Delta T_{adiabatic} = \frac{3R}{2}.59T = 7.4T Joules[/tex]

    [tex]Q = nC_v\Delta T_{isochoric} = \frac{3R}{2}(2.41T) = 30.0T Joules[/tex]

    AM
     
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