- #1

PhDeezNutz

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- Homework Statement
- Derive the condition the Adiabatic Condition ##PV^{\gamma} = \text{constant}## for a monatomic ideal gas. I believe ##\gamma = \frac{5}{3}##.

- Relevant Equations
- Equipartition Theorem:

##E = \frac{3}{2} NkT \Rightarrow \Delta E = \frac{3}{2} Nk \Delta T##

Ideal Gas Law:

## PV = NkT \Rightarrow P \Delta V + V \Delta P = Nk \Delta T ##

First Law:

##\Delta E = Q - W ##

Adiabatic:

##Q = 0##

By the First Law, Definition of an Adiabatic Process, and Definition of Work:

##\Delta E = Q - W = - W = - P \Delta V ## (because ##Q = 0##) (Equation 1)

By the Equipartition Theorem:

##\Delta E = \frac{3}{2} Nk \Delta T ## (Equation 2)

By Setting Equation 1 equal to Equation 2

## \Delta T = - \frac{2P}{3Nk} \Delta V## (Equation 3)

Differential Form of Ideal Gas Law:

##P \Delta V + V \Delta P = Nk \Delta T## (Equation 4)

Plug Equation 3 into Equation 4

## P \Delta V + V \Delta P = Nk \Delta T = - \frac{2P}{3} dV## (Equation 5)

So getting rid of the middle man

## P \Delta V + V \Delta P = - \frac{2P}{3} dV## (Equation 6)

Moving the LHS over to the RHS we have

##\frac{5}{3} P \Delta V + V \Delta P = 0## (Equation 7)

This can be stated another way

## \frac{\Delta P}{\Delta V} + \frac{5}{3} \left( \frac{1}{V} \right) P = 0## (Equation 8)

Equation 8 can be re-written in a calculus friendly way and we can use some basic methods of Differential Equations to solve for ##P## as a function of ##V## or just stop a step before doing that and establish ##PV^{\gamma} = \text{constant}##

## \frac{ dP}{dV} + \frac{5}{3} \left( \frac{1}{V} \right) P = 0## (Equation 8)

multiply through by integrating factor ##\mu = e^{\int \frac{5}{3V} \,dV} = e^{ln v^{\frac{5}{3}}} = v^{\frac{5}{3}}##

## v^{\frac{5}{3}} \frac{ dP}{dV} + \frac{5}{3} \left(v^{\frac{2}{3}} \right) P = 0## (Equation 8)

We recognize the LHS as the derivative of ##PV^{\frac{5}{3}}## So

##\frac{d}{dV} \left( PV^{\frac{5}{3}}\right) = 0## (Equation 9)

So

##PV^{\frac{5}{3}} = \text{constant}## (Equation 10)

of course if we really wanted to we could say

##P = \frac{\text{constant}}{V^{\frac{5}{3}}}##

Hopefully I didn't play too fast and loose with derivatives ##d## and full changes ##\Delta##, thanks in advance for any help/guidance.

If the above work is correct, how can we establish that "Adiabatic process on the PV diagram are steeper than those of isotherms".............after all we don't necessarily know what each "constant" is in

##PV = \text{constant}_1 ## (Isotherm)

##PV^{\frac{5}{3}} = \text{constant}_2## (Adiabat)

Are ##\text{constant}_1## and ##\text{constant}_2## the same?

##\Delta E = Q - W = - W = - P \Delta V ## (because ##Q = 0##) (Equation 1)

By the Equipartition Theorem:

##\Delta E = \frac{3}{2} Nk \Delta T ## (Equation 2)

By Setting Equation 1 equal to Equation 2

## \Delta T = - \frac{2P}{3Nk} \Delta V## (Equation 3)

Differential Form of Ideal Gas Law:

##P \Delta V + V \Delta P = Nk \Delta T## (Equation 4)

Plug Equation 3 into Equation 4

## P \Delta V + V \Delta P = Nk \Delta T = - \frac{2P}{3} dV## (Equation 5)

So getting rid of the middle man

## P \Delta V + V \Delta P = - \frac{2P}{3} dV## (Equation 6)

Moving the LHS over to the RHS we have

##\frac{5}{3} P \Delta V + V \Delta P = 0## (Equation 7)

This can be stated another way

## \frac{\Delta P}{\Delta V} + \frac{5}{3} \left( \frac{1}{V} \right) P = 0## (Equation 8)

Equation 8 can be re-written in a calculus friendly way and we can use some basic methods of Differential Equations to solve for ##P## as a function of ##V## or just stop a step before doing that and establish ##PV^{\gamma} = \text{constant}##

## \frac{ dP}{dV} + \frac{5}{3} \left( \frac{1}{V} \right) P = 0## (Equation 8)

multiply through by integrating factor ##\mu = e^{\int \frac{5}{3V} \,dV} = e^{ln v^{\frac{5}{3}}} = v^{\frac{5}{3}}##

## v^{\frac{5}{3}} \frac{ dP}{dV} + \frac{5}{3} \left(v^{\frac{2}{3}} \right) P = 0## (Equation 8)

We recognize the LHS as the derivative of ##PV^{\frac{5}{3}}## So

##\frac{d}{dV} \left( PV^{\frac{5}{3}}\right) = 0## (Equation 9)

So

##PV^{\frac{5}{3}} = \text{constant}## (Equation 10)

of course if we really wanted to we could say

##P = \frac{\text{constant}}{V^{\frac{5}{3}}}##

Hopefully I didn't play too fast and loose with derivatives ##d## and full changes ##\Delta##, thanks in advance for any help/guidance.

If the above work is correct, how can we establish that "Adiabatic process on the PV diagram are steeper than those of isotherms".............after all we don't necessarily know what each "constant" is in

##PV = \text{constant}_1 ## (Isotherm)

##PV^{\frac{5}{3}} = \text{constant}_2## (Adiabat)

Are ##\text{constant}_1## and ##\text{constant}_2## the same?

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