Thermodynamic constant -- misunderstanding

[mohamed_a]

I was reading about thermodynamics in my textbook wheni came across the following thermodynamics constants:

However, i don't understand why did we define 1/V inthe constants. What is the point in doing this?

 

[anuttarasammyak]

By dividing by V, physical dimensions or units of dotted $\alpha,\beta$ become simple, i.e. $T^{-1},P^{-1}.$
Now dotted $\alpha,\beta$ do not depend on volume of the system e.g. 200 ml or 400 ml of volume prepared in the experiments do not matter for measurement of these constants.

 

[Chestermiller]

$\dot{\alpha}$ and $\dot{\beta}$ are not constants. They vary (typically gradually) with both temperature and pressure. However, the V's are included in these definitions because $\dot{\alpha}$ is what we conventionally define as the coefficient of volumetric thermal expansion and $\dot{\beta}$ is what we conventionally define as the bulk compliance of a material (the reciprocal of the bulk modulus).

 

[mohamed_a]

By dividing by V, physical dimensions or units of dotted $\alpha,\beta$ become simple, i.e. $T^{-1},P^{-1}.$
Now dotted $\alpha,\beta$ do not depend on volume of the system e.g. 200 ml or 400 ml of volume prepared in the experiments do not matter for measurement of these constants.

i still can't find a use of this .So, omitting the V would just make the coefficient more intuitive, for example alpha being meter cube/ kelvin this points more to a rate which makes more sense.

 

[Chestermiller]

Why would your method be better than giving the % change in volume per unit change in temperature? Besides, your method would depend on the initial volume, and this definition wouldn't. Plus, their definition gives a value that is much more constant than yours does. Do you really think you are smarter than these brilliant scientists who worked this out and studied it over the past few hundred years?

 

[mohamed_a]

$\dot{\alpha}$ and $\dot{\beta}$ are not constants. They vary (typically gradually) with both temperature and pressure. However, the V's are included in these definitions because $\dot{\alpha}$ is what we conventionally define as the coefficient of volumetric thermal expansion and $\dot{\beta}$ is what we conventionally define as the bulk compliance of a material (the reciprocal of the bulk modulus).

So is it just a matter of definition?

Why would your method be better than giving the % change in volume per unit change in temperature? Besides, your method would depend on the initial volume, and this definition wouldn't. Plus, their definition gives a value that is much more constant than yours does. Do you really think you are smarter than these brilliant scientists who worked this out and studied it over the past few hundred years?

that's a probing question not an objection because i couldn't grasp the intuition. However, i understood it when i read wikipedia's page about it. the problem is i didn't understand its meaning because i didn't apply it on an example.

 

[anuttarasammyak]

i still can't find a use of this .So, omitting the V would just make the coefficient more intuitive, for example alpha being meter cube/ kelvin this points more to a rate which makes more sense.

I would tell more about this implication
of definition
$$\alpha(T,p)=(\frac{\partial \log \frac{V}{V_0}}{\partial T})_p$$
where $V_0$ is volume with temperature $T=T_0$
For simplicity of notion under the condition p=const. through the discussion
$$\alpha(T)=\frac{d \log \frac{V}{V_0}}{d T}$$
It is integrated to be
$$V=V_0 e^{\int_{T_0}^T \alpha(T)dt}$$
In case $\alpha$ is constant
$$V=V_0 e^{\alpha (T-T_0)}$$
Further when $\alpha (T-T_0) << 1$
$$V=V_0 (1+\alpha (T-T_0))$$
We can make use of thus defined $\alpha$ to express thermal expansion nature of matter in such a way. I hope you would share its convenience with us.

 

[mohamed_a]

I would tell more about this implication
of definition
$$\alpha(T,p)=(\frac{\partial \log \frac{V}{V_0}}{\partial T})_p$$
where $V_0$ is volume with temperature $T=T_0$
For simplicity of notion under the condition p=const. through the discussion
$$\alpha(T)=\frac{d \log \frac{V}{V_0}}{d T}$$
It is integrated to be
$$V=V_0 e^{\int_{T_0}^T \alpha(T)dt}$$
In case $\alpha$ is constant
$$V=V_0 e^{\alpha (T-T_0)}$$
Further when $\alpha (T-T_0) << 1$
$$V=V_0 (1+\alpha (T-T_0))$$
We can make use of thus defined $\alpha$ to express thermal expansion nature of matter in such a way. I hope you would share its convenience with us.

thanks for your generosity. the explanation is amazing and it deepened my understanding