Differential Integration Problem

[CGandC]

Homework Statement

I want to find the function $E(V,T)$

Relevant Equations

It is known that

$\frac{\partial E}{\partial T} = \frac{B N^2}{ V}$

$\frac{\partial E}{\partial V} = \frac{-B T N^2}{ V^2}$

$E(T=0) = 0$

Where B, N are constants.

Attempt at solution:

Writing the chain rule for $E(V,T)$:

$dE = \frac{\partial E}{\partial T}dT + \frac{\partial E}{\partial V}dV$

Then, integrating the differential:

$\int{ dE } = \int{ \frac{\partial E}{\partial T}dT } + \int{ \frac{\partial E}{\partial V}dV }$

If I put the partial derivatives:
$\int{ dE } = \int{\frac{B N^2}{ V}dT } + \int{ \frac{-B T N^2}{ V^2}dV }$
Then I get ( with using the condition $E(T=0) = 0$ ):
$E(T,V) = \frac{2BT N^2}{ V}$

Why my method of finding E(T,V) by integrating over all partial derivatives at once is incorrect( as opposed to the solution below)?

________________________________________________________________________

The right solution is:

$E(T,V) = \frac{BT N^2}{ V}$
Explanation for the right solution:

Integrating the following partial derivative:
$\frac{\partial E}{\partial V} = \frac{-B T N^2}{ V^2}$

I get:

$E(T,V) = \frac{B T N^2}{ V} + C(T)$

Using the condition $E(T=0) = 0$, I get that:
$C(T) = 0$
Therefore:
$E(T,V) = \frac{B T N^2}{ V}$

 

[pasmith]

Problem Statement: I want to find the function $E(V,T)$
Relevant Equations: It is known that

$\frac{\partial E}{\partial T} = \frac{B N^2}{ V}$

$\frac{\partial E}{\partial V} = \frac{-B T N^2}{ V^2}$

$E(T=0) = 0$

Where B, N are constants.

Attempt at solution:

Writing the chain rule for $E(V,T)$:

$dE = \frac{\partial E}{\partial T}dT + \frac{\partial E}{\partial V}dV$

Then, integrating the differential:

$\int{ dE } = \int{ \frac{\partial E}{\partial T}dT } + \int{ \frac{\partial E}{\partial V}dV }$

If I put the partial derivatives:
$\int{ dE } = \int{\frac{B N^2}{ V}dT } + \int{ \frac{-B T N^2}{ V^2}dV }$
Then I get ( with using the condition $E(T=0) = 0$ ):
$E(T,V) = \frac{2BT N^2}{ V}$

Why my method of finding E(T,V) by integrating over all partial derivatives at once is incorrect( as opposed to the solution below)?

Because in general$$E = \int \frac{\partial E}{\partial T}\,dT + F(V) = \int \frac{\partial E}{\partial V}\,dV + G(T).$$ What you are doing with your differential formulation is calculating $$\int \frac{\partial E}{\partial T}\,dT + \int \frac{\partial E}{\partial V}\,dV = 2E - F(V) - G(T).$$ As in this case $F = G = 0$ this is giving you twice the correct answer.

 

[CGandC]

Because in general$$E = \int \frac{\partial E}{\partial T}\,dT + F(V) = \int \frac{\partial E}{\partial V}\,dV + G(T).$$ What you are doing with your differential formulation is calculating $$\int \frac{\partial E}{\partial T}\,dT + \int \frac{\partial E}{\partial V}\,dV = 2E - F(V) - G(T).$$ As in this case $F = G = 0$ this is giving you twice the correct answer.

So instead of writing:
$dE = \frac{\partial E}{\partial T}dT + \frac{\partial E}{\partial V}dV$Why from the beginning we don't write:
$2dE = \frac{\partial E}{\partial T}dT + \frac{\partial E}{\partial V}dV$
in order to fix the mistake?

 

[vela]

Because it's wrong. You're suggesting making another mistake to compensate for your other mistake. Generally, that's not a good idea.

 

[CGandC]

Because it's wrong. You're suggesting making another mistake to compensate for your other mistake. Generally, that's not a good idea.

Why is it wrong?