1. May 12, 2010

### vorcil

I solved for the equation

$$\frac{f}{2} ln\frac{Tf}{Ti} = - ln \frac{vf}{vi}$$

I know to simplify the equation I need to Exponentiate both sides

$$e^{\frac{f}{2}} * e^{ln\frac{Tf}{Ti}} = e^{ln\frac{vi}{vf}}$$

note i changed $$- ln \frac{vf}{vi} \textrm{ too } ln\frac{vi}{vf}$$ and removed the negative sign

,

after exponentiating both sides I get

$$e^{\frac{f}{2}} * \frac{Tf}{Ti} = \frac{Vi}{Vf}$$

But I'm not sure if I put the f/2 in the right place

$${\frac{Tf}{Ti}}^\frac{f}{2}$$
but I don't know how to get to that
from exponentiating $$\frac{f}{2} \frac{Tf}{Ti}$$

could someone please expain the rules behind this?

--------------------------------------------

original question:

solve an equation for the work done in adiabatic compression

-

$$U = \frac{f}{2} N \kappa T$$

$$dU = \frac{f}{2} N \kappa dT$$

the work done during quasistatic compression is -P dv

$$\frac{f}{2} N\kappa dT = -P dV$$

using the ideal gas law to write P the pressure in terms of the variables T and v
ideal gas law: $$\frac{N \kappa T}{V} = P$$

i get

$$\frac{f}{2} N\kappa dT = -\frac{N\kappa T}{V} dv$$

simplifying by canceling out N K gives

$$\frac{f}{2} dT = -\frac{T}{V}dv$$

separation of variables because this is a differential equation

$$\frac{f}{2} \frac{dT}{T} = -\frac{dV}{V}$$

integrating both sides with respect to initial and final volume/temperatures
I get the equation

$$\frac{f}{2} \int_{Ti}^{Tf} \frac{1}{T}dT = -\int_{Vi}^{Vf} \frac{1}{V}dV$$

integrating

$$\left \frac{f}{2} lnT \right|_{Ti}^{Tf}= \left -lnV\right|_{Vi}^{Vf}$$

evaluating

$$\frac{f}{2} lnTf - lnTi = -lnVf + lnVi$$

noting that lnx - lny is equivalent to ln(x/y)

simplifying to get

$$\frac{f}{2} ln\frac{Tf}{Ti} = ln\frac{Vi}{Vf}$$

this is where I was having the exponential problem

exponentiating both sides

do i raise e to the whole of the left side? or the individual parts?
$$e^\frac{f}{2} e^{ln\frac{Tf}{Ti}} \textrm{ or do i } e^{\frac{f}{2} ln{Tf}{Ti}$$
what happens to the f/2?
I know each of Tf and Ti should be raised to the power of f/2 afterwards

but I do not understand how that works

2. May 12, 2010

### Je m'appelle

$$x^ax^b = x^{(a+b)}$$

$$\frac{x^a}{x^b} = x^{(a-b)}$$

$$(x^a)^b = x^{ab}$$

So, when you exponentiate

$$\frac{f}{2} ln\frac{Tf}{Ti} = ln\frac{Vi}{Vf}$$

You get

$$e^{\frac{f}{2} ln\frac{T_f}{T_i}} = e^{ln\frac{V_i}{V_f}}$$

But remember that

$$\frac{f}{2} ln\frac{T_f}{T_i} = ln(\frac{T_f}{T_i})^{(\frac{f}{2})}$$

Plug-in that on your equation to get

$$(\frac{T_f}{T_i})^{(\frac{f}{2})} = \frac{V_i}{V_f}$$

Was that what you wanted?

________________________

EDIT:

$$e^{\frac{f}{2}}e^{ln\frac{T_f}{T_i}} = e^{ln\frac{V_i}{V_f}}$$

On the left-hand side, you would be implying that

$$e^{\frac{f}{2}}e^{ln\frac{T_f}{T_i}} = e^{(\frac{f}{2}+ln\frac{T_f}{T_i})}$$

Which is incorrect, as you can clearly see, because the original expression was

$$\frac{f}{2}ln\frac{T_f}{T_i}$$

And not

$$\frac{f}{2}+ln\frac{T_f}{T_i}$$

So a quick explanation to your question would be reviewing the first exponential property that I've posted above, that is, the product one.

Last edited: May 12, 2010