Thermodynamic Adiabats: Solving U=aP^2V

[Joe Cool]

Homework Statement

The internal Energy of a system is $U=aP^2V$ with a positive constant a. Find the adiabats of this system in the P-V plane.
The solution is $$P=\frac 1 a\left( \sqrt{\frac {V_0} V}-1\right)$$

2. The attempt at a solution
the first law with the given internal energy:
$$a(2PVdP+P^2dV)=-PdV$$
Integration:
$$\frac {2adP} {1+aP}=- \frac{dV} V$$
$$2 ln(1+aP) = -lnV+const.$$
and
$$(1+aP)^2=\frac 1 V+const.$$
$$P=\frac 1 a\left( \sqrt{\frac 1 V+const}-1\right)$$

 

[TSny]

$$2 ln(1+aP) = -lnV+const.$$
and
$$(1+aP)^2=\frac 1 V+const.$$

You have a small but crucial error in going from the first equation above to the second.

 

[Joe Cool]

Thanks for your Tip. Should it be like that?
$ln((1+aP)^2)=ln\frac 1 V+lnV_0=ln\frac {V_0} V$ with $V_0=e^{const.}$

 

[TSny]

Thanks for your Tip. Should it be like that?
$ln((1+aP)^2)=ln\frac 1 V+lnV_0=ln\frac {V_0} V$ with $V_0=e^{const.}$

Yes, that's a nice way to do it.

The important thing is that if you have something like $\ln y = \ln x + const$, it does not follow that $y = x + const$.

If you have $\ln y = \ln x + k$ where $k$ is a constant, then "taking exponentials of both sides" yields
$e^{\ln y} = e^{\ln x + k} = e^k e^{\ln x}$.

So, $\ln y = \ln x + k$ implies $y = C x$ where $C = e^k$ is a new constant.