Recognitions:
Homework Help

 Quote by DrDu So setting $x=V_2/V_1$ and $y=P_2/P_1$ the first equation reads $$y_\mathrm{it}=x^{-1}$$ and the second one $$y_\mathrm{ad}=x^{-\gamma}$$.
What are V1, V2, P1 and P2? Why would x be the same for the adiabat and the isotherm?

AM

Recognitions:
 Quote by Andrew Mason What are V1, V2, P1 and P2? Why would x be the same for the adiabat and the isotherm? AM
The ordinate in the OP's graph is V_2, the abscissa is P_2, the two curves intersect at V_1, P_1.
Hence x will be the same for both graphs as it is just the scaled value of the ordinate.
 Recognitions: Gold Member 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.

Recognitions:
Homework Help
 Quote by DrDu The ordinate in the OP's graph is V_2, the abscissa is P_2, the two curves intersect at V_1, P_1. Hence x will be the same for both graphs as it is just the scaled value of the ordinate.
Ok. V2 is just a value of V other than V1 and V1 is the value of V where the graphs intersect.

You could also derive this by applying the adiabatic condition at the point where the slopes are equal and using T = γTiso

$$T_{ad}V^{\gamma-1} = T_{iso}V_1^{\gamma-1}$$ (adiabatic condition)

$$T_{ad} = T_{iso}\left(\frac{V_1}{V}\right)^{\gamma-1} = T_{iso}x^{1-\gamma}$$ where x = V/V1

and since:

$$T_{ad} = \frac{T_{iso}}{\gamma}$$

then:

$$x^{1-\gamma} = \frac{1}{\gamma} --> x = \gamma^{\frac{1}{\gamma-1}}$$

AM
 Recognitions: Gold Member 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.

Recognitions:
Homework Help
 Quote by Chestermiller 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.
The question asked about the slopes of the isothermal and adiabatic curves on a PV graph not a LogP-LogV graph.

AM

Recognitions:
Gold Member
 Quote by Andrew Mason The question asked about the slopes of the isothermal and adiabatic curves on a PV graph not a LogP-LogV graph. AM
No. The question didn't mention the word slopes. The question was, how you can prove mathematically that "the adiabatic curve is steeper than the isothermal curve." It didn't say what kind of plot you had to use to prove this (although, in all fairness, I must concede that the OP did display an arithmetic plot). There is really a question in semantics here as to what one means by the term steeper. In my judgment, for a given % change in volume, if the % change in pressure is greater for an adiabatic expansion than for an isothermal expansion (under all conditions), then the adiabatic variation is steeper. The log-log approach is consistent with working with percentage changes, and greatly simplifies the comparison. However, it all boils down to semantics and personal taste.
 $C_{p} = C_{V} + νRT, γ = \frac{C_{p}}{C_{V}} \geq 1.$ Hence the expression: Adiabatic: $p = const./V^{γ}$ Isothermal: p = const./V when γ is greater than 1, p has a steeper curve.
 @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.

Recognitions:
Homework Help