Internal energy, enthelpy and heat capacity

[MexChemE]

Hello PF! I have some questions regarding these concepts. First of all, are the following expressions valid for any case? i.e. any kind of process, like isochoric, isobaric.
$$\Delta U = \int C_v \ dT$$
$$\Delta H = \int C_p \ dT$$
Or is the ΔU expression only valid when dV = 0, and ΔH when dP = 0?

Also, when we are having an irreversible process, we have a gas with pressure P and it's going to expand against an opposing pressure p. When do we use the specific heat at constant pressure C_p, when the gas pressure P is constant, or when the opposing pressure p is constant? I'm also having the same doubt with enthalpy, which pressure are we using in the function of enthalpy? Is it H = U + pV or H = U + PV?

Thanks in advance!

 

[Chestermiller]

Hello PF! I have some questions regarding these concepts. First of all, are the following expressions valid for any case? i.e. any kind of process, like isochoric, isobaric.
$$\Delta U = \int C_v \ dT$$
$$\Delta H = \int C_p \ dT$$
Or is the ΔU expression only valid when dV = 0, and ΔH when dP = 0?

These expressions are valid for all ideal gases. You need to think of U and H as physical properties of the material, independent of the process that takes you from equilibrium state A to equilibrium state B.

For all materials, the heat capacities are defined by:
$$C_v=\left(\frac{\partial U}{\partial T}\right)_V$$$$C_p=\left(\frac{\partial H}{\partial T}\right)_P$$

Also, when we are having an irreversible process, we have a gas with pressure P and it's going to expand against an opposing pressure p.

In an irreversible process, the gas pressure P will not necessarily be uniform spatially within the system (so P is not well defined), but, at the interface with the surroundings, P and p match. So the work done on the surroundings is pdV (or $P_{interface}dV$).

When do we use the specific heat at constant pressure C_p, when the gas pressure P is constant, or when the opposing pressure p is constant?

C_p applies only to equilibrium states, so it is always P. But you can't use C_p along an irreversible path. It only applies to the end points (equilibrium states), and, you can get the change in enthalpy between the end points by integrating C_p with respect to T (for an ideal gas).

Is it H = U + pV or H = U + PV?

At the initial and final equilibrium states, there is no difference between these. For intermediate states along an irreversible path, neither expression applies. You can only use these at the equilibrium end points of the process.

Note that when you solve a problem involving an irreversible process at constant p, you do not consider the intermediate states of the system (except in integrating pdV to get the work), but only the two (equilibrium) end points. Look back in your notes and see if this is not correct.

Chet

 

[dipole]

They're only true when $V$ and $P$ are held constant, repsectively. In general, writing $U = U(T,V)$ for example,$$\Delta U = \int dU = \int \left(\frac{\partial U}{\partial T}\right)_VdT + \int \left(\frac{\partial U}{\partial V}\right)_TdV$$

and

$$\Delta U = \int C_VdT + \int \left(\frac{\partial U}{\partial V}\right)_TdV$$

and so if $V$ is held constant, the second term vanishes.

 

[Chestermiller]

They're only true when $V$ and $P$ are held constant, repsectively.

They're true for the case of an ideal gas even when V and P are not held constant, respectively, because $(\partial U/\partial V)_T$ and $(\partial H/\partial P)_T$ are both zero for ideal gases.

Chet

 

[MexChemE]

Thanks, Chet! Let's see if I got it correctly, if a gas did work against a constant pressure, then the process is isobaric and ΔH = Q, correct? If we heat the gas in an isochoric process, there was no work done, but we still have to account for the change of pressure in the change of enthalpy, right? I mean, ΔH = ΔU + VΔP?

Thank you both!

 

[Chestermiller]

Thanks, Chet! Let's see if I got it correctly, if a gas did work against a constant pressure, then the process is isobaric and ΔH = Q, correct?

Yes.

If we heat the gas in an isochoric process, there was no work done, but we still have to account for the change of pressure in the change of enthalpy, right?

Yes, if you want to know the change in enthalpy, although, for an ideal gas, you can also get it from the integral of CpdT (irrespective of the pressure change).

I mean, ΔH = ΔU + VΔP?

Yes. Also, for an ideal gas VΔP=RΔT.

Chet

 

[MexChemE]

Thanks! It feels nice when you finally begin to understand how basic concepts are applied to many different situations. By the way, is your blog about the first two laws of thermodynamics still available?

 

[Chestermiller]

Thanks! It feels nice when you finally begin to understand how basic concepts are applied to many different situations. By the way, is your blog about the first two laws of thermodynamics still available?

I'll send you a copy in a conversation message.

Chet

 

[Chestermiller]

I wasn't able to do it in a conversation, so here it is:

FIRST LAW OF THERMODYNAMICS

Suppose that we have a closed system that at initial time t_i

is in an initial equilibrium state, with internal energy U_i

, and at a later time t_f, it is in a new equilibrium

state with internal energy U_f. The transition from the

initial equilibrium state to the final equilibrium state is brought about

by imposing a time-dependent heat flow across the interface between the

system and the surroundings, and a time-dependent rate of doing work at

the interface between the system and the surroundings. Let $\dot{q} (t)$ represent the rate of heat addition across the interface

between the system and the surroundings at time t, and let $\dot{w} (t)$ represent the rate at which the system does work on the

surroundings at the interface at time t. According to the first law

(basically conservation of energy),
$$\Delta U=U_f-U_i=\int_{t_i}^{t_f}{(\dot{q}(t)-\dot{w}(t))dt}=Q-W$$
where Q is the total amount of heat added and W is the total amount of

work done by the system on the surroundings at the interface.

The time variation of $\dot{q}(t)$ and $\dot{w}(t)$ between the initial and final states uniquely characterizes the

so-called process path. There are an infinite number of possible process

paths that can take the system from the initial to the final equilibrium

state. The only constraint is that Q-W must be the same for all of them.

If a process path is irreversible, then the temperature and pressure

within the system are inhomogeneous (i.e., non-uniform, varying with

spatial position), and one cannot define a unique pressure or temperature

for the system (except at the initial and the final equilibrium state).

However, the pressure and temperature at the interface can be

measured and controlled using the surroundings to impose the temperature

and pressure boundary conditions that we desire. Thus, T_I(t)

and P_I(t) can be used to impose the process path that we

desire. Alternately, and even more fundamentally, we can directly

control, by well established methods, the rate of heat flow and the rate

of doing work at the interface $\dot{q}(t)$ and $\dot {w}(t)$).

Both for reversible and irreversible process paths, the rate at

which the system does work on the surroundings is given by:
$$\dot{w}(t)=P_I(t)\dot{V}(t)$$
where $\dot{V}(t)$ is the rate of change of system volume at

time t. However, if the process path is reversible, the pressure P

within the system is uniform, and

$P_I(t)=P(t)$ (reversible process path)

Therefore, $\dot{w}(t)=P(t)\dot{V}(t)$ (reversible process

path)

Another feature of reversible process paths is that they are carried out

very slowly, so that $\dot{q}(t)$ and $\dot{w}(t)$

are both very close to zero over then entire process path. However, the

amount of time between the initial equilibrium state and the final

equilibrium state (t_f-t_i) becomes exceedingly

large. In this way, Q-W remains constant and finite.

SECOND LAW OF THERMODYNAMICS

In the previous section, we focused on the infinite number of process

paths that are capable of taking a closed thermodynamic system from an

initial equilibrium state to a final equilibrium state. Each of these

process paths is uniquely determined by specifying the heat transfer

rate $\dot{q}(t)$ and the rate of doing work $\dot{w}(t)$ as functions of time at the interface between the system and the

surroundings. We noted that the cumulative amount of heat transfer and

the cumulative amount of work done over an entire process path are given

by the two integrals:
$$Q=\int_{t_i}^{t_f}{\dot{q}(t)dt}$$
$$W=\int_{t_i}^{t_f}{\dot{w}(t)dt}$$
In the present section, we will be introducing a third integral of this

type (involving the heat transfer rate $\dot{q}(t)$) to

provide a basis for establishing a precise mathematical statement of the

Second Law of Thermodynamics.

The discovery of the Second Law came about in the 19th century, and

involved contributions by many brilliant scientists. There have been

many statements of the Second Law over the years, couched in complicated

language and multi-word sentences, typically involving heat reservoirs,

Carnot engines, and the like. These statements have been a source of

unending confusion for students of thermodynamics for over a hundred

years. What has been sorely needed is a precise mathematical definition

of the Second Law that avoids all the complicated rhetoric. The sad part

about all this is that such a precise definition has existed all along.

The definition was formulated by Clausius back in the 1800's.

Clausius wondered what would happen if he evaluated the following

integral over each of the possible process paths between the initial and

final equilibrium states of a closed system:
$$I=\int_{t_i}^{t_f}{\frac{\dot{q}(t)}{T_I(t)}dt}$$
where T_I(t) is the temperature at the interface with the

surroundings at time t. He carried out extensive calculations on many

systems undergoing a variety of both reversible and irreversible paths

and discovered something astonishing. He found that, for any closed

system, the values calculated for the integral over all the possible

reversible and irreversible paths (between the initial and final

equilibrium states) was not arbitrary; instead, there was a unique upper

bound (maximum) to the value of the integral. Clausius also found that

this result was consistent with all the "word definitions" of the Second

Law.

Clearly, if there was an upper bound for this integral, this upper bound

had to depend only on the two equilibrium states, and not on the path

between them. It must therefore be regarded as a point function of

state. Clausius named this point function Entropy.

But how could the value of this point function be determined without

evaluating the integral over every possible process path between the

initial and final equilibrium states to find the maximum? Clausius made

another discovery. He determined that, out of the infinite number of

possible process paths, there existed a well-defined subset, each member

of which gave the same maximum value for the integral. This subset

consisted of what we call today the reversible process paths.

So, to determine the change in entropy between two equilibrium states,

one must first conceive of a reversible path between the states and then

evaluate the integral. Any other process path will give a value for the

integral lower than the entropy change.

So, mathematically, we can now state the Second Law as follows:

$$I=\int_{t_i}^{t_f}{\frac{\dot{q}(t)}{T_I(t)}dt}\leq\Delta S=\int_ {t_i}^{t_f} {\frac{\dot{q}_{rev}(t)}{T(t)}dt}$$
where $\dot{q}_{rev}(t)$ is the heat transfer rate for any of

the reversible paths between the initial and final equilibrium states,

and T(t) is the system temperature at time t (which, for a

reversible path, is equal to the temperature at the interface with the

surroundings). This constitutes a precise mathematical statement of the

Second Law of Thermodynamics.