# Thermodynamics and polytropic ideal gas (very simple theory question)

1. Jul 7, 2013

### joshmccraney

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!

2. Jul 7, 2013

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.

3. Jul 7, 2013

### Andrew Mason

The P is missing a subscript.

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

So:

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

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

AM

4. Jul 7, 2013

### joshmccraney

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?

5. Jul 7, 2013

### joshmccraney

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