Homework Help: Question involving integration

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 pv^k = p₀v₀^k.

2. The attempt at a solution

[tex]\frac{dp}{dv} = \frac{p}{v}[/tex]
[tex]\frac{p}{dp} = \frac{v}{dv}[/tex]
[tex]\int \frac{p}{dp} = \int \frac{v}{dv}[/tex]
[tex]\ln |p| = \ln |v| + \bar{c}[/tex]
[tex]e^{\ln v + \bar{c}} = p[/tex]
[itex]p = Cv^k[/itex] where [itex]C = e^{\bar{c}}[/itex]
[itex]p_0 = Cv_0^k[/itex]
So, it appears that p0/v0^k = p/v^k = C, but I'm not sure how to obtain that pv^k = p₀v₀^k.

Thank you in advance.

 

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:
[tex]\frac{dp}{dv} = k\frac{p}{v}[/tex]
[tex]\frac{dp}{p} = k\frac{dv}{v}[/tex]
[tex]\ln|p| = k\ln|v| + \bar{c}[/tex]
[tex]p = e^{k\ln v + \bar{c}}[/tex]
[tex]p = Cv^k[/tex]
But I'm still not sure how to obtain the desired result.

[EDIT]
If I write
[tex]\frac{dp}{p} = -k\frac{dv}{v}[/tex]
with a minus sign in front of k, I obtain:
[tex]\ln|p| = -k\ln|v| + \bar{c}[/tex]
[tex]\ln|p| + k\ln|v| = \bar{c}[/tex]
[tex]\ln|p| + \ln|v|^k = \bar{c}[/tex]
[tex]\ln|p||v|^k = \bar{c}[/tex]
[tex]pv^k = e^{\bar{c}}[/tex]
[tex]pv^k = C[/tex]
which seems to be the desired result, but I'm not very sure about it.

Thank you in advance.

 

Very good then, thank you for confirming.