Entropy and Heat Capacity Relation

In summary, the equation you are trying to solve is ill-defined because it only has two independent variables. You can solve it using the heat capacities and the thermodynamic potentials.
  • #1
cwill53
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I have a simple question sort of about exact differentials and deciding which variables matter and when.

I know we can write entropy ##S## as ##S(P,T)## and ##S(V,T)## to derive different relations between heat capacities ##C_V## and ##C_P##. I was wondering if it is technically correct to write

$$dS(P,V,T)=\left ( \frac{\partial S}{\partial P} \right )_{V,T}dP+\left ( \frac{\partial S}{\partial V} \right )_{P,T}dV+\left ( \frac{\partial S}{\partial T} \right )_{P,V}dT$$

$$\delta Q_{rev}=TdS(P,V,T)=T\left ( \frac{\partial S}{\partial P} \right )_{V,T}dP+T\left ( \frac{\partial S}{\partial V} \right )_{P,T}dV+T\left ( \frac{\partial S}{\partial T} \right )_{P,V}dT$$

Then how would we define heat capacities ##C,C_V,C_P## in terms of these partial derivatives?

I know we can define them as (at constant volume and constant pressure, respectively)

$$C_V=T\left ( \frac{\partial S}{\partial T} \right )_{V}; C_P=T\left ( \frac{\partial S}{\partial T} \right )_{P}$$

But I can't understand how to reconcile the gap between the second and third equations I typed.
 
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  • #2
There's a couple points of confusion here.

First, the first equation you wrote is not correct. Those partial derivatives are not independent. A given thermodynamic system has state variables P,V,T, and N. For a closed system, N is fixed. That means you're down to 3 remaining variables, as in your equation. However, there's always another constraint: the equation of state (e.g., ideal gas law or van der waals gas law), which reflects the unique physics of the system. This brings you down to two independent variables. In other words, any set of three state variables (P,V,T) or (P,S,T) or what have you, is over-defined.

Second, what is this third heat capacity C (as opposed to ##C_v## or ##C_p##) that you mention? The other two contain everything you need. Remember, any reversible process can be broken down into and isobaric or isochoric part plus an isothermal part. Irreversible processes that are internally reversible can be handled by the same approach if the rate of irreversibility is known.
 
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  • #3
Twigg said:
There's a couple points of confusion here.

First, the first equation you wrote is not correct. Those partial derivatives are not independent. A given thermodynamic system has state variables P,V,T, and N. For a closed system, N is fixed. That means you're down to 3 remaining variables, as in your equation. However, there's always another constraint: the equation of state (e.g., ideal gas law or van der waals gas law), which reflects the unique physics of the system. This brings you down to two independent variables. In other words, any set of three state variables (P,V,T) or (P,S,T) or what have you, is over-defined.

Second, what is this third heat capacity C (as opposed to ##C_v## or ##C_p##) that you mention? The other two contain everything you need. Remember, any reversible process can be broken down into and isobaric or isochoric part plus an isothermal part. Irreversible processes that are internally reversible can be handled by the same approach if the rate of irreversibility is known.
I ended up finding out that the phase space isn’t three dimensional; and also that ##S(P,V,T)## is illegal anyway as the third variable is already determined by the other two, and also S is not a thermodynamic potential. I can derive the heat capacities using ##S(P,T)## and ##S(V,T)## easily, I was just confused on how ##S(P,V,T)## was ill-defined. I understand now though.
 
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Related to Entropy and Heat Capacity Relation

1. What is entropy and how does it relate to heat capacity?

Entropy is a measure of the disorder or randomness of a system. Heat capacity, on the other hand, is a measure of the amount of heat energy required to raise the temperature of a substance. The two are related because as the disorder or randomness of a system increases, the heat capacity also increases.

2. How does entropy impact the behavior of a system?

Entropy plays a crucial role in determining the behavior of a system. As entropy increases, the system becomes more disordered and less stable. This can lead to changes in temperature, pressure, and other physical properties of the system.

3. How is entropy calculated?

The entropy of a system can be calculated using the formula S = k ln(W), where k is the Boltzmann constant and W is the number of microstates (possible arrangements of particles) in the system. This formula is based on statistical mechanics and is used to describe the relationship between entropy and the probability of a system being in a particular state.

4. What factors affect the heat capacity of a substance?

The heat capacity of a substance is affected by several factors, including the mass, temperature, and molecular structure of the substance. It also depends on the conditions under which the substance is heated, such as constant pressure or constant volume.

5. How does the relationship between entropy and heat capacity impact thermodynamic processes?

The relationship between entropy and heat capacity is fundamental to understanding thermodynamic processes. It helps to explain why some processes are spontaneous and others are not, and how energy is transferred and transformed in a system. It also plays a role in determining the efficiency of a process and the maximum amount of work that can be extracted from a system.

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