Is My Homework Solution Correct?

[Pushoam]

The derivation goes like this:

$dH=dU+PdV+VdP=\left(\frac{\partial U}{\partial T}\right)_VdT+\left(\frac{\partial U}{\partial V}\right)_TdV+PdV+VdP$ ... (1)

In the above eqn., U is taken as a function of V and T. So, is it correct to assume V and T as independent variables?

In the second eqn.,

$\left(\frac{\partial H}{\partial T}\right)_P=\left(\frac{\partial U}{\partial T}\right)_V+\left(\frac{\partial U}{\partial V}\right)_T\left(\frac{\partial V}{\partial T}\right)_P+P\left(\frac{\partial V}{\partial T}\right)_P$....(2)

We calculate $\frac{\partial V}{\partial T}$. This means that V is a function of two variables T and P.
Then, $\left( \frac{\partial U}{\partial V}\right)_ T = -P + T\left( \frac{\partial P}{\partial T}\right)_V$
This means that p is a function of V and T.

So, the independent variable is only T. V is an implicit function of P and T. P is an implicit function of V and T.
U is a function of V and T.
Is this correct?

 

[Chestermiller]

In the above eqn., U is taken as a function of V and T. So, is it correct to assume V and T as independent variables?

Yes. U can be expressed as a function of V and T.

In the second eqn.,

We calculate $\frac{\partial V}{\partial T}$. This means that V is a function of two variables T and P.

Yes. This is the equation of state.

Then, $\left( \frac{\partial U}{\partial V}\right)_ T = -P + T\left( \frac{\partial P}{\partial T}\right)_V$
This means that p is a function of V and T.

Yes. The latter is the equation of state also.

So, the independent variable is only T.

I don't understand what this is supposed to mean. We are trying to develop the relationship between the partial derivative of H with respect to T at constant P, and the partial derivative of U with respect to T at constant V.

V is an implicit function of P and T. P is an implicit function of V and T.
U is a function of V and T.
Is this correct?

V being a function of P and T is equivalent to P being a function of V and T. Both of these represent the equation of state. U is usually expressed as a function of V and T, although it can also be expressed as a function of P and T, or also a function of P and V.

 

[Pushoam]

What I have understood is :

The state of the system could be characterised by P, V and T.

We have a equation of state relating P,V and T.

Any two of {P,V,T} could act as independent variables depending upon our experiment and the 3rd would be the dependent variable.

U could be a function of any of the two independent variables depending upon the experiment.

Here, we have taken U as a function of V and T.

Now, the equation:

$C_P – C_V = {\left [ {\left (\left ( \frac {\partial U } {\partial V } \right) _{T} +P \right) } \left ( \frac {\partial V } {\partial T } \right) _{P} \right ] }$ ……….(1)$dQ = \left ( \frac {\partial U } {\partial T } \right) _{V} dT + \left ( \frac {\partial U } {\partial V } \right) _{T} dV + p dV$ …….(2)

In the above equation Q is a function of V and T.

$\left ( \frac {\partial Q } {\partial T } \right) _{V} = \left ( \frac {\partial U } {\partial T } \right) _{V} + 0 +0 = C_V$....(3)

The more correct formulation is:

Q itself is a physical quantity. Depending of the experiment, sometimes it could be a function of V and T or P and T i.e. whether Q is Q = Q(V,T) or Q = Q(P,T) depends upon the experiment.

When Q is a function of V and T i.e. Q = Q( V,T),

We have,

$\left ( \frac {\partial Q } {\partial T } \right) _{V} = \left ( \frac {\partial U } {\partial T } \right) _{V} + 0 +0 = C_V$....(3)

When Q is a function of P and T i.e. Q = Q( P,T),

We have,

$\left ( \frac {\partial Q } {\partial T } \right) _{P} = C_P= \left ( \frac {\partial U } {\partial T } \right) _{V} + \{ \left ( \frac {\partial U } {\partial V } \right) _{T} + p\} \left ( \frac {\partial V } {\partial T } \right) _{P}$ …….(4)

In equation 2 and 3, we have expressed both U and Q as a function of V and T.

In equation 4, we have expressed U as a function of V and T( so we have got equation 2), but we have expressed Q as a function of P and T.

Isn’t it always true that when U is a function of V and T, then Q must be a function of V and T or when Q is a function of P and T, then U must be a function of P and T?

In equation 4, we have expressed Q = Q (P,T), so the independent variables are P and T.

V is a function of P and T i.e. V = V(P,T), while U is a function of V and T i.e. U = U(V,T). Since here V is a function of P and T, U = U(V,T) = U(V(P,T), T) = U(P,T).

So, expressing U as a function of P and T,

We have,

$C_P = \left ( \frac {\partial Q } {\partial T } \right) _{P} = \left ( \frac {\partial U } {\partial T } \right) _{P} +P \left ( \frac {\partial V } {\partial T } \right) _{P}$

Then, we have to find $\left ( \frac {\partial U } {\partial T } \right) _{P}$ applying Maxwell relations upon G(P,T).

Is this correct?

 

[Pushoam]

In the following equation

dU = dQ + dW, where dW is the work done on the system.

We write, dW = -p dV, where p is the external pressure acting on the system.

While writing dW = -p dV, is it assumed that p is constant, i.e. dW = - p dV – V dp

But because of constant p , dW = -pdV

 

[Chestermiller]

What I have understood is :

The state of the system could be characterised by P, V and T.

We have a equation of state relating P,V and T.

Any two of {P,V,T} could act as independent variables depending upon our experiment and the 3rd would be the dependent variable.

U could be a function of any of the two independent variables depending upon the experiment.

Here, we have taken U as a function of V and T.

Now, the equation:

$C_P – C_V = {\left [ {\left (\left ( \frac {\partial U } {\partial V } \right) _{T} +P \right) } \left ( \frac {\partial V } {\partial T } \right) _{P} \right ] }$ ……….(1)$dQ = \left ( \frac {\partial U } {\partial T } \right) _{V} dT + \left ( \frac {\partial U } {\partial V } \right) _{T} dV + p dV$ …….(2)

In the above equation Q is a function of V and T.

$\left ( \frac {\partial Q } {\partial T } \right) _{V} = \left ( \frac {\partial U } {\partial T } \right) _{V} + 0 +0 = C_V$....(3)

The more correct formulation is:

Q itself is a physical quantity. Depending of the experiment, sometimes it could be a function of V and T or P and T i.e. whether Q is Q = Q(V,T) or Q = Q(P,T) depends upon the experiment.

When Q is a function of V and T i.e. Q = Q( V,T),

We have,

$\left ( \frac {\partial Q } {\partial T } \right) _{V} = \left ( \frac {\partial U } {\partial T } \right) _{V} + 0 +0 = C_V$....(3)

When Q is a function of P and T i.e. Q = Q( P,T),

We have,

$\left ( \frac {\partial Q } {\partial T } \right) _{P} = C_P= \left ( \frac {\partial U } {\partial T } \right) _{V} + \{ \left ( \frac {\partial U } {\partial V } \right) _{T} + p\} \left ( \frac {\partial V } {\partial T } \right) _{P}$ …….(4)

In equation 2 and 3, we have expressed both U and Q as a function of V and T.

In equation 4, we have expressed U as a function of V and T( so we have got equation 2), but we have expressed Q as a function of P and T.

Isn’t it always true that when U is a function of V and T, then Q must be a function of V and T or when Q is a function of P and T, then U must be a function of P and T?

In equation 4, we have expressed Q = Q (P,T), so the independent variables are P and T.

V is a function of P and T i.e. V = V(P,T), while U is a function of V and T i.e. U = U(V,T). Since here V is a function of P and T, U = U(V,T) = U(V(P,T), T) = U(P,T).

So, expressing U as a function of P and T,

We have,

$C_P = \left ( \frac {\partial Q } {\partial T } \right) _{P} = \left ( \frac {\partial U } {\partial T } \right) _{P} +P \left ( \frac {\partial V } {\partial T } \right) _{P}$

Then, we have to find $\left ( \frac {\partial U } {\partial T } \right) _{P}$ applying Maxwell relations upon G(P,T).

Is this correct?

I am very confused by what you have done here. Are you aware that, in irreversible processes, P, T, or even both P and T are non-uniform within the system. In that case, what value of P or T do you use? If you are calling Q the enthalpy, or is it the heat? If it is the heat, then, even for reversible processes between two thermodynamic equilibrium states, Q and W are not unique. So, you can't express Q as a function of the parameters P, V, and T at a thermodynamic equilibrium state (even if you did also include the parameters at the initial state).

 

[Pushoam]

As Q (heat)is not a state function, we cannot express it as a function of state variables.
But then doesn't the eqn.
$dQ = \left ( \frac {\partial U } {\partial T } \right) _{V} dT + \left ( \frac {\partial U } {\partial V } \right) _{T} dV + p dV$ tell that Q is a function of V and T?
$\left ( \frac {\partial Q } {\partial T } \right) _{V} =C_V$ means that Q is fn of V and T, dosn't it?
A partial differentiation is defined only for a well defined function. If Q is not a fn of V and T, then Q is a fn of which variables?
If Q is not a well - defined fn, then how do we calculate heat given to a system for a given process?

Actually, Q is well - defined. The point is : the value of Q is different for different process. Q is a process dependent fn or path dependent fn. So, when the path is given, Q is defined. For $\left ( \frac {\partial Q } {\partial T } \right) _{V} =C_V$ , the process is an isochoric process, and hence, for this process, Q is a function of T. Is this correct?

I myself am getting confused here. So sorry if it doesn't make sense what I am writing here. I will take another try. I may be then able to formulate better where I am stuck.

 

[Charles Link]

As Q (heat)is not a state function, we cannot express it as a function of state variables.
But then doesn't the eqn.
$dQ = \left ( \frac {\partial U } {\partial T } \right) _{V} dT + \left ( \frac {\partial U } {\partial V } \right) _{T} dV + p dV$ tell that Q is a function of V and T.
$\left ( \frac {\partial Q } {\partial T } \right) _{V} =C_V$ means that Q is fn of V and T.
A partial differentiation is defined only for a well defined function. If Q is not a fn of V and T, then Q is a fn of which variable?

I myself am getting confused here. So sorry if it doesn't make what I am writing here. I will take another try. I may be then able to formulate where I am stuck.

I think I can help answer this question: Generally $Q$ is not used in expressions like these. If it is a reversible process, then what is used in place of $dQ$ is $TdS$. The entropy $S$ is a state function. Wikipedia calls $Q$ and $W$ "process functions" as compared to the other quantities such as internal energy $U$, enthalpy $H$, and entropy $S$, which are "state functions".

 

[Pushoam]

I think I can help answer this question: Generally $Q$ is not used in expressions like these. If it is a reversible process, then what is used in place of $dQ$ is $TdS$. The entropy $S$ is a state function. Wikipedia calls $Q$ and $W$ "process functions" as compared to the other quantities such as internal energy $U$, enthalpy $H$, and entropy $S$, which are "state functions".

dQ = T dS for a reversible process.
Since both S and T are state - fn.,T dS is same for each path. Hence, the if we integrate TdS between the final and initial states, the integral i.e. Q also should be independent of the path. Does this mean that Q is a path - independent function for a reversible process?
But in thermodynamics, mostly, reversible processes are studied as each step of reversible process is in thermodynamic equillibrium and only in this eqbm, the state variables are well - defined.

 

[Charles Link]

dQ = T dS for a reversible process.
Since both S and T are state - fn.,T dS is same for each path. Hence, the if we integrate TdS between the final and initial states, the integral i.e. Q also should be independent of the path. Does this mean that Q is a path - independent function for a reversible process?
But in thermodynamics, mostly, reversible processes are studied as each step of reversible process is in thermodynamic equillibrium and only in this eqbm, the state variables are well - defined.

I think what you are saying is correct, but @Chestermiller is really the Thermodynamics expert. It takes a lot of practice to get proficient at working thermodynamic problems. The entropy $S$ is a very different concept that takes a while to become well-acquainted with it . $\\$ In a reversible process, at each step in the process, I believe there is equilibrium. You can pause a reversible process in the middle of the process=e.g. the reversible expansion of a gas, and the system will remain in that equilibrium state. And what you say is also correct, in irreversible processes, the state functions are not well defined, and not unique.

 

[Chestermiller]

As Q (heat)is not a state function, we cannot express it as a function of state variables.
But then doesn't the eqn.
$dQ = \left ( \frac {\partial U } {\partial T } \right) _{V} dT + \left ( \frac {\partial U } {\partial V } \right) _{T} dV + p dV$ tell that Q is a function of V and T?

This equation is not valid for an irreversible path. That is because, unless the path is reversible, the work is not pdV. Moreover, it is always a bad idea to write a thermodynamic equation in terms of differentials unless the path is reversible. Even then, this equation would be valid only for the change between two closely neighboring equilibrium states. Plus, if you took two different tortuous reversible paths between the same two closely neighboring equilibrium states, in general, the total Q and the total W could differ between paths; they would just be nearly equal to one another for each path.

$\left ( \frac {\partial Q } {\partial T } \right) _{V} =C_V$ means that Q is fn of V and T, dosn't it?

No. The correct relationship is $\left ( \frac {\partial U } {\partial T } \right) _{V} =C_V$. U is a unique function of V and T for equilibrium states, not Q.

A partial differentiation is defined only for a well defined function. If Q is not a fn of V and T, then Q is a fn of which variables?
If Q is not a well - defined fn, then how do we calculate heat given to a system for a given process?

Q is the amount of heat added from the surroundings to the system, irrespective of whether the process is reversible or irreversible? Check your textbook and see how Q is calculated for various processes. Sometimes it is specified in the problem statement, sometimes it is backed out of the first law (knowing the work and the change in internal energy), and sometimes it is determine from the change in the surroundings internal energy. It is never calculated from a change in V and T.

Actually, Q is well - defined. The point is : the value of Q is different for different process. Q is a process dependent fn or path dependent fn. So, when the path is given, Q is defined. For $\left ( \frac {\partial Q } {\partial T } \right) _{V} =C_V$ , the process is an isochoric process, and hence, for this process, Q is a function of T. Is this correct?

This is one of the cases in which is it backed out from the first law $\Delta U$.

 

[Chestermiller]

dQ = T dS for a reversible process.
Since both S and T are state - fn.,T dS is same for each path. Hence, the if we integrate TdS between the final and initial states, the integral i.e. Q also should be independent of the path. Does this mean that Q is a path - independent function for a reversible process?

This is not correct mathematically. We can create many Carnot cycles in which the temperatures at which heat transfer occurs are the same and the change in entropy is the same (zero), but for which the net heat transferred is not the same.

 

[Pushoam]

This equation is not valid for an irreversible path. That is because, unless the path is reversible, the work is not pdV.

The above equation comes from the 1st law of thermodynamics ( $\Delta U = \Delta Q + \Delta W$)which is valid for ( I think ) both reversible and irreversible processes.

For reversible process, U as a function of V and T and dW = - p dV are defined. So, the above equation is valid only for reversible process. Right ?

Even then, this equation would be valid only for the change between two closely neighboring equilibrium states.

Is this because the above eqn deals with the infinitesimal change and the difference between two closely neighboring eqbm states is infinitesimal?

$\left ( \frac {\partial Q } {\partial T } \right) _{V} =C_V$means that Q is fn of V and T, dosn't it?

No. The correct relationship is $\left ( \frac {\partial U } {\partial T } \right) _{V} =C_V$. U is a unique function of V and T for equilibrium states, not Q.

The book says,

Here, $C_V$ is defined as $C_V = \left ( \frac {\partial Q } {\partial T } \right) _{V}$ and then on calculation it is found that $C_V = \left ( \frac {\partial U } {\partial T } \right) _{V}$.

What I have understood from your statement as $C_V$ is defined as $C_V = \left ( \frac {\partial U } {\partial T } \right) _{V}$.
Is this correct?

Q is the amount of heat added from the surroundings to the system, irrespective of whether the process is reversible or irreversible? Check your textbook and see how Q is calculated for various processes. Sometimes it is specified in the problem statement, sometimes it is backed out of the first law (knowing the work and the change in internal energy), and sometimes it is determine from the change in the surroundings internal energy. It is never calculated from a change in V and T.

For any reversible process,

$\Delta U$ between any two states or dU between any two close states is well defined independently of path.

$\Delta Q$ between any two states or dQ between any two close states is well defined for any given path.

So, U is well defined as a function of state variables independently of the path.

And Q is well defined as a function of the state variables for a given path. Isn’t it so?

The book says that for reversible process both TdS and pdV are differrential of state fn. So, their definite integral i.e Q and W respectively should also be state fn. But, it is said that Q and W are not state fn. This is where I am stuck.

 

[Chestermiller]

View attachment 214876
The above equation comes from the 1st law of thermodynamics ( $\Delta U = \Delta Q + \Delta W$)which is valid for ( I think ) both reversible and irreversible processes.

For reversible process, U as a function of V and T and dW = - p dV are defined. So, the above equation is valid only for reversible process. Right ?

For an irreversible path, p and T are not uniform within the system, and viscous stresses contribute to the force per unit area that the gas exerts on the surroundings, so the equation of state can not be used to establish the force on the surroundings or the work. Also, for both an irreversible path or a reversible path, the "volumetric" work done by the system on the surroundings can be determined by integrating $p_{ext}dV$, where $p_{ext}$ is the force per unit area exerted by the surroundings on the system (if this force is somehow known or specified). In the case of a reversible path, $p_{ext}$ also matches the pressure of the gas within the system p calculated from the equation of state.

Is this because the above eqn deals with the infinitesimal change and the difference between two closely neighboring eqbm states is infinitesimal?

Yes.

The book says,
View attachment 214876

Here, $C_V$ is defined as $C_V = \left ( \frac {\partial Q } {\partial T } \right) _{V}$ and then on calculation it is found that $C_V = \left ( \frac {\partial U } {\partial T } \right) _{V}$.

What I have understood from your statement as $C_V$ is defined as $C_V = \left ( \frac {\partial U } {\partial T } \right) _{V}$.
Is this correct?

I completely disagree with how this is presented in your book, and, as justification for this disagreement, I cite your present state of confusion.

In freshman physics, they taught us that $Q=C\Delta T$, where they called C the heat capacity of the material and Q the heat transferred to the material. All the problems to which this was applied were ones in which no work was done.

Now we arrive at thermodynamics, and we encounter situations where work can be done on the surroundings. And immediately we encounter problems with using this old definition because Q is supposed to be (exclusively) the heat transferred from the surroundings to the system, while C is supposed to be a physical property of the material. And, when work is being done, Q is not longer equal to $C\Delta T$. So, in thermodynamics, we find it necessary to discard the old definition of C in terms of Q, and adopt a new definition in which we consistently get the right answer, whether or not work is being done, while, at the same time reducing to the old definition when work is not being done. This new definition is:
$$C_v=\left(\frac{\partial U}{\partial T}\right)_V$$and $$C_p=\left(\frac{\partial H}{\partial T}\right)_P$$These new definitions clear up all the difficulties with Q being energy in transit (specifically for a process) while C is a physical property of the material (independent of any process).

For any reversible process,

$\Delta U$ between any two states or dU between any two close states is well defined independently of path.

$\Delta Q$ between any two states or dQ between any two close states is well defined for any given path.

There are an infinite number of $\Delta Q$s or dQs between any two close states for the infinite number of paths between the two states. But there is only one $\Delta U$ or dU. So, clearly, Q is not a unique function of state. It is a function only of path. And the path cannot be specified giving the system temperature and pressure as a function of time along the path, because these are not uniform within the system for an irreversible path.

So, U is well defined as a function of state variables independently of the path.

Yes.

And Q is well defined as a function of the state variables for a given path. Isn’t it so?

No.

View attachment 214880
The book says that for reversible process both TdS and pdV are differrential of state fn. So, their definite integral i.e Q and W respectively should also be state fn. But, it is said that Q and W are not state fn. This is where I am stuck.

Q is not the definite integral of TdS and W is not the definite integral of PdV. This is only true for a reversible path. For an irreversible process, we can only know T at the boundary of the system (typically an ideal reservoir temperature, $T_R$), and, for such a process, the most that we can say is that $$\int{\frac{dQ}{T_R}}\lt\Delta S$$where $\Delta S$ is the change in entropy between the initial and final thermodynamic equilibrium states of the system (i.e., assuming that the final state is an equilibrium state). And, for both reversible and irreversible processes, $$W=\int{p_{ext}dV}$$.

Here is an article I wrote for Physics Forums Insights that I wrote a couple of years ago that might help: https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/