Thermodynamics and polytropic ideal gas (very simple theory question)

[member 428835]

hey all!

i have a question i was hoping some of you could unravel. specifically, in thermodynamics i understand in a quasi-static situation we can right work as:

W=\int PdV where W is work, P is pressure, and V is volume.

my book defines polytropic to be PV^n = constant

it then writes the following polytropic, quasi-static equality when n=1 , which is where i am lost:

W=\int PdV = \int P \frac{{V_1}}{V} dV = P{V_1} \int \frac{{dV}}{V} = P{V_1} ln(\frac{{V_2}}{V_1})

specifically, if PV_1 is constant, then if we pull it out of the integral how is it we integrate over \frac{1}{V} ? why doesn't the equality implode here: \int P \frac{{V_1}}{V} dV = P{V_1} \int \frac{{dV}}{V} if V_1 is a constant why isn't V? any help is greatly appreciated!

 

[boneh3ad]

I think either you copied something wrong our your book is leaving out a small detail here. Namely, for the polytropic process with n=1, you have the relation simplified to
pV = c.
So, if you have some base state that represents the starting conditions of your system, you can say
p_1 V_1 = c.
There, p_1 and V_1 are both constants because they simply represent the pressure and volume of your system at its initial state. You can rearrange this for substitution into the work equation,
p = \dfrac{c}{V} = \dfrac{p_1 V_1}{V}
They are initial conditions. Going back to your work equation,
W = \int\limits_1^2 p\;dV = \int\limits_1^2 \dfrac{p_1 V_1}{V}\;dV = p_1 V_1 \int\limits_1^2 \dfrac{1}{V}\;dV = p_1 V_1 \left.\ln(V)\right|_1^2 = p_1 V_1 \left[\ln(V_2) - \ln(V_1)\right] = p_1 V_1 \ln\left(\dfrac{V_2}{V_1}\right)
which recovers the answer in your question with the exception that there is a subscript 1 in front of the pressure term. The key is that you know that pV is equal to a constant, and you "know" the initial state of the system, so you can replace that constant with the values of p and V at that initial state, and both of those are constants and can be removed from the integral. In short, p_1 and [/itex]V_1[/itex] are constants while p and V are not because the former refer to specific values of the later at a specific time, and those don't change.

 

[Andrew Mason]

hey all!

i have a question i was hoping some of you could unravel. specifically, in thermodynamics i understand in a quasi-static situation we can right work as:

W=\int PdV where W is work, P is pressure, and V is volume.

my book defines polytropic to be PV^n = constant

it then writes the following polytropic, quasi-static equality when n=1 , which is where i am lost:

W=\int PdV = \int P \frac{{V_1}}{V} dV = P{V_1} \int \frac{{dV}}{V} = P{V_1} ln(\frac{{V_2}}{V_1})

The P is missing a subscript.

If PV = constant then PV must equal P₁V₁, so P = P₁V₁/V.

So:

W=\int PdV = \int P_1 \frac{{V_1}}{V} dV = P_1V_1 \int \frac{{dV}}{V} = P_1V_1 ln(\frac{{V_2}}{V_1})

specifically, if PV_1 is constant,

PV = constant. So PV₁ cannot be constant (ie. P cannot be constant) as V changes.

AM

 

[member 428835]

In short, p_1 and V_1 are constants while p and V are not because the former refer to specific values of the later at a specific time, and those don't change.

if this is true, how can we make the questionable integral equality? it seems if V_1 \neq V then we cannot have the equality that was pointed out. can you explain?

 

[member 428835]

never mind, it just "clicked". thanks guys! yay for PF saving the day again!