Thermodynamics: integrating partial derivatives

[psid]

Homework Statement

Given that mechanical equation of state for a paramagnetic substance is
$$m=\left(\frac{DH}{T}\right)$$
where D is a constant, H is the magnetic field, m is molar magnetization and the molar heat capacity
$$c_{m}$$ is constant, find entropy and enthalpy

Homework Equations

$$\left(\frac{\partial S}{\partial T}\right)_{m}=\frac{C_{m}}{T}$$
$$\left(\frac{\partial S}{\partial m}\right)_{T}=-\left(\frac{\partial H}{\partial T}\right)_{m}$$

The Attempt at a Solution

Integrating above equations I found that
$$S=-\frac{m^{2}}{2D}+c_{m}\ln{\frac{T}{T_{0}}}.$$
Or alternatively
$$T=T_{0}\exp{(2DS+m^{2})/(2DC)}.$$
Now to the entalphy h. By definition
$$\left(\frac{\partial h}{\partial S}\right)_{H}=T=T_{0}\exp{(2DS+m^{2})/(2DC)}$$
and
$$\left(\frac{\partial h}{\partial H}\right)_{S}=-m=DH/T.$$
Now, I'm a bit confused here. When integrating these kind of partial derivatives, I guess that one should express the integrand using only those variables that are used to perform the differentiation, i.e. S and H in this particular case. If this is the case, one should probably eliminate m from above equation using the equation of state. But this gives
$$T=T_{0}\exp{(2DS+DH^{2}/(2CT^{2}))/(2DC)}$$
which makes no sense as T appears on both sides... What's going wrong?

 

[anathema]

Thank you for your post. I would like to clarify a few things about your attempt at solving this problem.

Firstly, the equations you have listed as "homework equations" are not correct. The correct equations for entropy and enthalpy are:

\left(\frac{\partial S}{\partial T}\right)_{p}=\frac{C_{p}}{T}
\left(\frac{\partial S}{\partial p}\right)_{T}=\frac{V}{T}
\left(\frac{\partial h}{\partial T}\right)_{p}=C_{p}
\left(\frac{\partial h}{\partial p}\right)_{T}=V

where p is pressure, V is volume, and C_p is the molar heat capacity at constant pressure. These equations can be found in any thermodynamics textbook.

Secondly, the equation of state you have listed is not for a paramagnetic substance, but for an ideal gas. The equation of state for a paramagnetic substance is:

m=\frac{C_{m}H}{T}

where C_m is the molar heat capacity at constant magnetization. This equation can also be found in thermodynamics textbooks.

Finally, to solve for entropy and enthalpy, you need to use the correct equations and integrate appropriately. You cannot use the equations you have listed and expect to get the correct answers. I suggest consulting a thermodynamics textbook or seeking guidance from a professor or tutor.

 

[m2003]

It is important to note that in thermodynamics, the equations and variables used are often simplified and may not always perfectly represent real-world systems. In this case, the mechanical equation of state for a paramagnetic substance may not accurately reflect the behavior of a real substance. That being said, let's continue with the solution.

To find the entropy and enthalpy, we can use the given equations and integrate with respect to the appropriate variables. Starting with the first equation, we can integrate with respect to T and m to find the entropy:

\int\left(\frac{\partial S}{\partial T}\right)_{m}dT=\int\frac{C_{m}}{T}dT
\Rightarrow S=C_{m}\ln{T}+f(m)

Next, we integrate with respect to m:

\int\left(\frac{\partial S}{\partial m}\right)_{T}dm=\int-\left(\frac{\partial H}{\partial T}\right)_{m}dm
\Rightarrow S=-\frac{m^{2}}{2D}+g(T)

Combining these two equations, we can eliminate the arbitrary functions f(m) and g(T) to find the final expression for entropy:

S=-\frac{m^{2}}{2D}+C_{m}\ln{\frac{T}{T_{0}}}

Similarly, for enthalpy, we integrate the equations given in the problem:

\int\left(\frac{\partial h}{\partial S}\right)_{H}dS=\int TdS
\Rightarrow h=TS+f(H)

\int\left(\frac{\partial h}{\partial H}\right)_{S}dH=\int-mdH
\Rightarrow h=-\frac{m^{2}}{2D}+g(S)

Eliminating the arbitrary functions f(H) and g(S), we can find the final expression for enthalpy:

h=TS-\frac{m^{2}}{2D}

It is important to note that these equations may not accurately represent the behavior of a real substance and should be used with caution. Additionally, it is always important to check the units and consistency of the equations and variables used in any thermodynamic calculations.