Gas undergoing isothermic compression

[Karol]

Does a curved line in the P-V diagram for an adiabatic process have only a single temperature on each point of the line or can it have many?
In post #16 you showed that the temperature is -69 for work against pressure of 1[atm], and if we choose a different pressure, but with the same initial and final volumes, there will be different temperatures but the line will be the same since the P-V are the same, no? the lines follow $PV^{\gamma}=Const$
Is it possible to have different temperatures for a single point? the temperature is determined by PV=nRT and it has one value at each point (for the same amount of material n)

 

[Chestermiller]

Does a curved line in the P-V diagram for an adiabatic process have only a single temperature on each point of the line or can it have many?
In post #16 you showed that the temperature is -69 for work against pressure of 1[atm], and if we choose a different pressure, but with the same initial and final volumes, there will be different temperatures but the line will be the same since the P-V are the same, no? the lines follow $PV^{\gamma}=Const$
Is it possible to have different temperatures for a single point? the temperature is determined by PV=nRT and it has one value at each point (for the same amount of material n)

This is all true for a reversible process, where the gas passes through a continuous sequence of thermodynamic equilibrium states. Under these circumstances, the pressure, temperature, and molar density within the gas are constant with spatial position, and are related by the ideal gas law.

But in an irreversible expansion, where the gas expands very rapidly, the local temperature, pressure, and molar density within the gas vary with spatial position from one parcel of gas to another at any given time during the expansion. Only in the final equilibrium state do the temperature, pressure, and molar density become uniform again. But, what about during the expansion process? What value of the pressure do you use to calculate the work done on the surroundings if the pressure is not the same at all locations within the gas? (And, even if it were, how would you calculate the pressure using the ideal gas law if the temperature is varying with spatial position within the gas in some unknown way?) Well, to get the work done on the surroundings, you use the pressure at the interface with the surroundings, which you can manually measure and/or impose on the gas. That's your 1 atm. in this problem.

When blogs were still a feature of Physics Forums, I had a blog that explains all this in detail. When the blogs were removed, I saved my blog in a file in case I ever wanted to use it. Here is write up that I saved:

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 dream up 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.

 

[Karol]

This is all true for a reversible process, where the gas passes through a continuous sequence of thermodynamic equilibrium states. Under these circumstances, the pressure, temperature, and molar density within the gas are constant with spatial position, and are related by the ideal gas law.

Do you refer to $PV^{\gamma}=Const$? so in the case of resistance of 1[atm] and irreversible processes the graph doesn't follow $PV^{\gamma}=Const$? is this formula derived from the ideal gas law?
And in the case of 1[atm] resistance you cannot deduce anything except using the first law, as you did in post #16?

 

[Chestermiller]

Do you refer to $PV^{\gamma}=Const$?

Yes.

so in the case of resistance of 1[atm] and irreversible processes the graph doesn't follow $PV^{\gamma}=Const$?

No, it doesn't. For an irreversible process, you can only plot the pressure at the piston face P_I vs V, and it doesn't follow that equation.

is this formula derived from the ideal gas law?

It is derived exclusively for the case of an ideal gas undergoing an adiabatic reversible process (and nothing else).

And in the case of 1[atm] resistance you cannot deduce anything except using the first law, as you did in post #16?

You can still get the change in entropy (a 2nd law quantity) between the initial and final states, because it is a function of state only, and not of path.

Chet

 

[Karol]

So for any 2 states, (P_i, V_i) and (P_f, V_f) of an adiabatic process there is only one value of work done and it can be calculated using:
$$U=\frac{3}{2}nRT$$
And the first law? (ΔU=W)

 

[Karol]

In Wikipedia, under Specific Heat: $dU=\delta Q-\delta W$
And if the work is by expanding the boundary: $dU=\delta Q-PdV$
So for constant volume:
$$\left(\frac{\vartheta U}{\vartheta T}\right)=\left(\frac{\vartheta Q}{\vartheta T}\right)=C_v$$
So U is only a function of T. But you wrote:
$$dU=C_vdT+\left(\frac{\partial U}{\partial V}\right)_TdV$$
So is the article in Wikipedia about ideal gas or not? i know your formula is for real gasses. how to explain the contradiction

 

[Chestermiller]

In Wikipedia, under Specific Heat: $dU=\delta Q-\delta W$
And if the work is by expanding the boundary: $dU=\delta Q-PdV$
So for constant volume:
$$\left(\frac{\vartheta U}{\vartheta T}\right)=\left(\frac{\vartheta Q}{\vartheta T}\right)=C_v$$
So U is only a function of T. But you wrote:
$$dU=C_vdT+\left(\frac{\partial U}{\partial V}\right)_TdV$$
So is the article in Wikipedia about ideal gas or not? i know your formula is for real gasses. how to explain the contradiction

For an ideal gas, $$\left(\frac{\partial U}{\partial V}\right)_T=0$$

Chet

 

[Chestermiller]

Actually, dQ is not an exact differential unless the process path is reversible. But, for a reversible process path in which both the temperature and the volume are changing,
$$dQ=C_vdT+T\left(\frac{\partial P}{\partial T}\right)_VdV$$
You will learn how to derive this relationship once you study the 2nd Law.

Chet

 

[Karol]

Good but from the formula $dU=\delta Q-PdV$ there are no more members at all added to $dU=C_vdT$ and in your formula there is

 

[Chestermiller]

Good but from the formula $dU=\delta Q-PdV$ there are no more members at all added to $dU=C_vdT$ and in your formula there is

Substitute the relationship in post #43 into your equation for dU. Then collect terms and substitute the ideal gas law into the coefficient of dV, and see what you get.

Chet

 

[Karol]

$$\left(\frac{\partial U}{\partial V}\right)_TdV=\left(T\left(\frac{\partial P}{\partial T}\right)_V-P\right)dV\rightarrow\left(\frac{\partial U}{\partial V}\right)_T=\left(T\left(\frac{\partial P}{\partial T}\right)_V-P\right)$$
Why is this equality? is it because we have dV at both ends so logically we understand that it has to be the energy change due to pressure change?
And why $\left(\frac{\partial U}{\partial V}\right)_T$ is at constant temperature? why not $\left(\frac{\partial U}{\partial V}\right)_P$, at constant pressure?
Are those things taught in the second law?

 

[Chestermiller]

$$\left(\frac{\partial U}{\partial V}\right)_TdV=\left(T\left(\frac{\partial P}{\partial T}\right)_V-P\right)dV\rightarrow\left(\frac{\partial U}{\partial V}\right)_T=\left(T\left(\frac{\partial P}{\partial T}\right)_V-P\right)$$
Why is this equality? is it because we have dV at both ends so logically we understand that it has to be the energy change due to pressure change?

No. This is the effect of the volume change on the internal energy at constant temperature, not the effect of the pressure change.

The first term in the parenthesis comes from dQ, not from dW. For a single phase ideal gas of constant composition undergoing a reversible change:

$dU=C_vdT$

$dQ=C_vdT+PdV$

$dW=PdV$

The fact that the internal energy of an ideal gas is a function only of temperature originated experimentally (by Joule), by doing tests on real gases at low pressures where they approach ideal gas behavior. It was found that, if the gas was expanded or compressed at constant temperature (i.e., the initial and final temperatures were equal, say by using a constant temperature reservoir), irrespective of whether the process was reversible or irreversible, the amount of heat added to the gas Q was virtually identical to the work done on the gas W. This meant that, when the gas volume was changed at constant temperature, its internal energy U did not change.

And why $\left(\frac{\partial U}{\partial V}\right)_T$ is at constant temperature? why not $\left(\frac{\partial U}{\partial V}\right)_P$, at constant pressure?

The internal energy is a function of state. For a single phase substance of constant composition, the state of the substance is determined by any two independent physical parameters. Examples are P and T, P and V, and V and T (there are others). (In what we've been doing, U is the internal energy per mole, and V is the molar volume). All these combinations of pairs of physical parameters are equivalent. In the analysis we have been doing, we have been using the combination T and V (so U = U(T,V)). (The most convenient combinations to work with are T and V, and T and P). The key is that you have to choose the two that you will be working with. We can, of course, also do the analysis using dT and dP, and the partial of U with respect to T at constant P, and the partial of U with respect to P at constant T.

Are those things taught in the second law?

The equation
$$\left(\frac{\partial U}{\partial V}\right)_T=\left(T\left(\frac{\partial P}{\partial T}\right)_V-P\right)$$
is derived using the second law.

Chet

 

[Karol]

The internal energy is a function of state. For a single phase substance of constant composition, the state of the substance is determined by any two independent physical parameters. Examples are P and T, P and V, and V and T (there are others).

So Q is a function of path? that is why i can't analyze it as a function of 2 variables, (V,P), (T,P)... like you did for U?

 

[Chestermiller]

So Q is a function of path? that is why i can't analyze it as a function of 2 variables, (V,P), (T,P)... like you did for U?

Q is a function of path, but that's not the only reason. Only for reversible paths can Q be evaluated in terms of variations of P and T. For irreverisble paths, P and T are not uniform within the system (they vary with spatial position), so there are no single unique values of these parameters to use in the calculation.

Chet

 

[Karol]

Thank you very much Chet