- #1
vorcil
- 398
- 0
I solved for the equation
[tex] \frac{f}{2} ln\frac{Tf}{Ti} = - ln \frac{vf}{vi} [/tex]
I know to simplify the equation I need to Exponentiate both sides
[tex]
e^{\frac{f}{2}} * e^{ln\frac{Tf}{Ti}} = e^{ln\frac{vi}{vf}} [/tex]
note i changed [tex] - ln \frac{vf}{vi} \textrm{ too } ln\frac{vi}{vf} [/tex] and removed the negative sign
,
after exponentiating both sides I get
[tex] e^{\frac{f}{2}} * \frac{Tf}{Ti} = \frac{Vi}{Vf} [/tex]
But I'm not sure if I put the f/2 in the right place
My equation should be reading
[tex] {\frac{Tf}{Ti}}^\frac{f}{2} [/tex]
but I don't know how to get to that
from exponentiating [tex] \frac{f}{2} \frac{Tf}{Ti} [/tex]
could someone please expain the rules behind this?
--------------------------------------------
original question:
solve an equation for the work done in adiabatic compression
-
[tex] U = \frac{f}{2} N \kappa T [/tex]
[tex] dU = \frac{f}{2} N \kappa dT [/tex]
the work done during quasistatic compression is -P dv
[tex] \frac{f}{2} N\kappa dT = -P dV [/tex]
using the ideal gas law to write P the pressure in terms of the variables T and v
ideal gas law: [tex] \frac{N \kappa T}{V} = P [/tex]
i get
[tex] \frac{f}{2} N\kappa dT = -\frac{N\kappa T}{V} dv [/tex]
simplifying by canceling out N K gives
[tex] \frac{f}{2} dT = -\frac{T}{V}dv [/tex]
separation of variables because this is a differential equation
[tex]
\frac{f}{2} \frac{dT}{T} = -\frac{dV}{V} [/tex]
integrating both sides with respect to initial and final volume/temperatures
I get the equation
[tex] \frac{f}{2} \int_{Ti}^{Tf} \frac{1}{T}dT = -\int_{Vi}^{Vf} \frac{1}{V}dV [/tex]
integrating
[tex] \left \frac{f}{2} lnT \right|_{Ti}^{Tf}= \left -lnV\right|_{Vi}^{Vf} [/tex]
evaluating
[tex] \frac{f}{2} lnTf - lnTi = -lnVf + lnVi [/tex]
noting that lnx - lny is equivalent to ln(x/y)
simplifying to get
[tex] \frac{f}{2} ln\frac{Tf}{Ti} = ln\frac{Vi}{Vf} [/tex]
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?
[tex] e^\frac{f}{2} e^{ln\frac{Tf}{Ti}} \textrm{ or do i } e^{\frac{f}{2} ln{Tf}{Ti} [/tex]
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
[tex] \frac{f}{2} ln\frac{Tf}{Ti} = - ln \frac{vf}{vi} [/tex]
I know to simplify the equation I need to Exponentiate both sides
[tex]
e^{\frac{f}{2}} * e^{ln\frac{Tf}{Ti}} = e^{ln\frac{vi}{vf}} [/tex]
note i changed [tex] - ln \frac{vf}{vi} \textrm{ too } ln\frac{vi}{vf} [/tex] and removed the negative sign
,
after exponentiating both sides I get
[tex] e^{\frac{f}{2}} * \frac{Tf}{Ti} = \frac{Vi}{Vf} [/tex]
But I'm not sure if I put the f/2 in the right place
My equation should be reading
[tex] {\frac{Tf}{Ti}}^\frac{f}{2} [/tex]
but I don't know how to get to that
from exponentiating [tex] \frac{f}{2} \frac{Tf}{Ti} [/tex]
could someone please expain the rules behind this?
--------------------------------------------
original question:
solve an equation for the work done in adiabatic compression
-
[tex] U = \frac{f}{2} N \kappa T [/tex]
[tex] dU = \frac{f}{2} N \kappa dT [/tex]
the work done during quasistatic compression is -P dv
[tex] \frac{f}{2} N\kappa dT = -P dV [/tex]
using the ideal gas law to write P the pressure in terms of the variables T and v
ideal gas law: [tex] \frac{N \kappa T}{V} = P [/tex]
i get
[tex] \frac{f}{2} N\kappa dT = -\frac{N\kappa T}{V} dv [/tex]
simplifying by canceling out N K gives
[tex] \frac{f}{2} dT = -\frac{T}{V}dv [/tex]
separation of variables because this is a differential equation
[tex]
\frac{f}{2} \frac{dT}{T} = -\frac{dV}{V} [/tex]
integrating both sides with respect to initial and final volume/temperatures
I get the equation
[tex] \frac{f}{2} \int_{Ti}^{Tf} \frac{1}{T}dT = -\int_{Vi}^{Vf} \frac{1}{V}dV [/tex]
integrating
[tex] \left \frac{f}{2} lnT \right|_{Ti}^{Tf}= \left -lnV\right|_{Vi}^{Vf} [/tex]
evaluating
[tex] \frac{f}{2} lnTf - lnTi = -lnVf + lnVi [/tex]
noting that lnx - lny is equivalent to ln(x/y)
simplifying to get
[tex] \frac{f}{2} ln\frac{Tf}{Ti} = ln\frac{Vi}{Vf} [/tex]
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?
[tex] e^\frac{f}{2} e^{ln\frac{Tf}{Ti}} \textrm{ or do i } e^{\frac{f}{2} ln{Tf}{Ti} [/tex]
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