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Isothermal vs Adiabatic expansion |
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| Sep25-12, 11:13 AM | #18 |
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Isothermal vs Adiabatic expansionAM |
| Sep25-12, 01:42 PM | #19 |
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Hence x will be the same for both graphs as it is just the scaled value of the ordinate. |
| Sep25-12, 04:50 PM | #20 |
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DrDu is correct.
The differential equation for any isotherm (for an ideal gas) is: d lnP/ d lnV =-1 If we integrate this equation, we get ln P = - ln V + CI (1) If P = P1 at V = V1, then CI = (ln P1 - ln V1) (2) The differential equation for any adiabat (for an ideal gas) is: d lnP/ d lnV =-γ If we integrate this equation, we get ln P = - γ ln V + CA (3) If P = P1 at V = V1, then CA = (ln P1 - γ ln V1) (4) Equations 1 and 3 are the equations for two straight lines on a log-log plot, and Eqns. 2 and 4 indicate that the two straight lines intersect at the single point (P1 , V1). Actually any given adiabat can only intersect any given isotherm at one and only one point in P-V space. Eqns. 1 and 3 indicate that the absolute magnitude of the slope of an isotherm on a log-log plot is constant, and less than the magnitude of the slope of the adiabat on the same plot (which is also constant). This is what "steeper" means to me. |
| Sep27-12, 12:51 AM | #21 |
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You could also derive this by applying the adiabatic condition at the point where the slopes are equal and using T = γTiso [tex]T_{ad}V^{\gamma-1} = T_{iso}V_1^{\gamma-1}[/tex] (adiabatic condition) [tex]T_{ad} = T_{iso}\left(\frac{V_1}{V}\right)^{\gamma-1} = T_{iso}x^{1-\gamma}[/tex] where x = V/V1 and since: [tex]T_{ad} = \frac{T_{iso}}{\gamma}[/tex] then: [tex]x^{1-\gamma} = \frac{1}{\gamma} --> x = \gamma^{\frac{1}{\gamma-1}}[/tex] AM |
| Sep27-12, 06:26 AM | #22 |
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On a log-log plot, the slopes are never equal. The slope of log P vs log V for adiabatic is always steeper than the corresponding slope for isothermal.
Also, I see no reason for bringing temperature back into the picture, since the temperature effect is inherently included in the derivation of the adiabatic equation through the parameter γ. |
| Sep27-12, 07:38 AM | #23 |
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| Sep27-12, 08:20 AM | #24 |
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| Sep27-12, 10:09 AM | #25 |
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[itex]C_{p} = C_{V} + νRT, γ = \frac{C_{p}}{C_{V}} \geq 1. [/itex]
Hence the expression: Adiabatic: [itex]p = const./V^{γ}[/itex] Isothermal: p = const./V when γ is greater than 1, p has a steeper curve. |
| Sep27-12, 10:26 AM | #26 |
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@raopeng
We have already been through this and you are making the same mistake I made originally. The state variables, P,V & T are only the same at one point on the graph - the intial point. At other points the temperature (and pressure) are different for the adiabat and isotherm at stated volume. |
| Sep28-12, 10:32 PM | #27 |
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| Sep29-12, 06:29 AM | #28 |
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