# Question involving integration

1. Nov 26, 2011

### pc2-brazil

1. The problem statement, all variables and given/known data
When a gas undergoes an adiabatic compression or expansion (it neither gains nor loses heat), then the rate of change of pressure, with respect to the volume, varies directly with the pressure and inversely with the volume. If the pressure is p N/m² when the volume is v m³ and the initial pressure and initial volume are p0 N/m² and v0 m³, respectively, show that pvk = p0v0k.

2. The attempt at a solution

$$\frac{dp}{dv} = \frac{p}{v}$$
$$\frac{p}{dp} = \frac{v}{dv}$$
$$\int \frac{p}{dp} = \int \frac{v}{dv}$$
$$\ln |p| = \ln |v| + \bar{c}$$
$$e^{\ln v + \bar{c}} = p$$
$p = Cv^k$ where $C = e^{\bar{c}}$
$p_0 = Cv_0^k$
So, it appears that p0/v0k = p/vk = C, but I'm not sure how to obtain that pvk = p0v0k.

2. Nov 26, 2011

### I like Serena

Where did you get that pressure varies inversely with volume?
I believe it doesn't in a reversible adiabatic process.

What you do have is:
du=-pdv (for a reversible adiabatic process)
du=cvdT
pv=RT (for an ideal gas)

If you combine them, I think you'll find a different relation for p and v.

I'm afraid this is not true.
You have p = C v.
There is no power k.

3. Nov 27, 2011

### pc2-brazil

I got that pressure varies inversely with volume from the question itself, unless I've misinterpreted it.

Yes, actually I wrote it wrongly. I forgot to add k, the proportionality constant, to the differential equation. It should be:
$$\frac{dp}{dv} = k\frac{p}{v}$$
$$\frac{dp}{p} = k\frac{dv}{v}$$
$$\ln|p| = k\ln|v| + \bar{c}$$
$$p = e^{k\ln v + \bar{c}}$$
$$p = Cv^k$$
But I'm still not sure how to obtain the desired result.

[EDIT]
If I write
$$\frac{dp}{p} = -k\frac{dv}{v}$$
$$\ln|p| = -k\ln|v| + \bar{c}$$
$$\ln|p| + k\ln|v| = \bar{c}$$
$$\ln|p| + \ln|v|^k = \bar{c}$$
$$\ln|p||v|^k = \bar{c}$$
$$pv^k = e^{\bar{c}}$$
$$pv^k = C$$
which seems to be the desired result, but I'm not very sure about it.

Last edited: Nov 27, 2011
4. Nov 27, 2011

### I like Serena

This is correct!

5. Nov 27, 2011

### pc2-brazil

Very good then, thank you for confirming.