# A few questions on an irreversible adiabatic expansion of an ideal gas

• Chemistry
• zenterix
zenterix
Homework Statement
Consider an irreversible adiabatic expansion of an ideal gas.
Relevant Equations
We start with an adiabatic container containing a gas at ##P_1, V_1##, and ##T_1##. There is a movable piston in the container. The external pressure is ##P_2<P_1##, and thus the piston is held in place by stoppers.
When we remove the stoppers, the gas expands and the piston shoots up and eventually reaches a new final position in which the internal and external pressures are the same.

Apparently we can write

$$\delta q=0\tag{1}$$

$$\delta w=-P_2dV\tag{2}$$

$$dU=C_VdT\tag{3}$$

$$dU=-P_2dV\tag{4}$$

I've studied the origin of all these equations but I just don't feel comfortable with them yet.

Thus, I have a few basic questions still.

##(1)## comes from the fact that the process is adiabatic.

I am confused about when we can or cannot write a differential form.

1) It seems that ##\delta w=-P_{ext}dV## can always be written no matter what (ie no matter if the process is reversible or irreversible). Is this so?

2) In a ##PV## diagram, we only know start and end points. According to a lecture I saw, we can't draw any sort of path connecting them because the process is irreversible.

We can depict the work done in this diagram. I just realized (if it is indeed correct) that when we do this we are not drawing a path but just graphically showing a quantity, the total work done. Is this correct?

3) ##U## is a state function. Suppse we start at a ##P_1,V_1##, and ##T_1## and move to a final ##P_2,V_2##. For our irreversible process we will have a ##T_2##.

For a reversible process we will have some ##T_{2,rev}## which may be different from ##T_2##. Thus, we aren't moving to the same final states necessarily.

(3) comes from writing out the differential form for ##U=U(T,V)## and assuming an ideal gas. Thus, for an ideal gas, (3) is "always" true. But is "always" really always here?

It seems not to matter that the irreversible process we have happens very suddenly and even violently. (3) expresses what happens when we change ##T## infinitesimally.

Reversibility just seems to imply being able to sub in the ideal gas law for ##P_2## in (2) and (4).

I suppose any very rapid process can be considered as a sequence of infinitesimal steps in time, but we are not talking about time in these equations, we are talking about temperature. Temperature does not seem to be necessarily the same throughout the system in this rapid process.

4) With regard to equation (4), I have similar doubts. I guess volume can be seen, like time, to increase infinitesimally if we just break the process into small enough steps. And external pressure ##P_2## is constant, so (4) seems to make sense. But then we're back to my question 1) above.

Last edited:
zenterix said:
Homework Statement: Consider an irreversible adiabatic expansion of an ideal gas.
Relevant Equations: We start with an adiabatic container containing a gas at ##P_1, V_1##, and ##T_1##. There is a movable piston in the container. The external pressure is ##P_2<P_1##, and thus the piston is held in place by stoppers.

When we remove the stoppers, the gas expands and the piston shoots up and eventually reaches a new final position in which the internal and external pressures are the same.

Apparently we can write

$$\delta q=0\tag{1}$$

$$\delta w=-P_2dV\tag{2}$$

$$dU=C_VdT\tag{3}$$

$$dU=-P_2dV\tag{4}$$

I've studied the origin of all these equations but I just don't feel comfortable with them yet.

Thus, I have a few basic questions still.

##(1)## comes from the fact that the process is adiabatic.

I am confused about when we can or cannot write a differential form.
I write differential forms of state energy functions (U,S,A,G) only when the process is reversible. Otherwise, only finite forms are to be used in the first- and 2nd laws of thermodynamics.
zenterix said:
1) It seems that ##\delta w=-P_{ext}dV## can always be written no matter what (ie no matter if the process is reversible or irreversible). Is this so?
Yes, but only for PV changes in a closed system.
zenterix said:
2) In a ##PV## diagram, we only know start and end points. According to a lecture I saw, we can't draw any sort of path connecting them because the process is irreversible.
If ##P_{ext}## is specified for the irreversible path, then it can be shown on a ##P_{ext}V## diagram.
zenterix said:
We can depict the work done in this diagram. I just realized (if it is indeed correct) that when we do this we are not drawing a path but just graphically showing a quantity, the total work done. Is this correct?
No, if ##P_{ext}## is specified as a function of V for the irreversible path, this is the path of the process.
zenterix said:
3) ##U## is a state function. Suppse we start at a ##P_1,V_1##, and ##T_1## and move to a final ##P_2,V_2##. For our irreversible process we will have a ##T_2##.
Yes. P2 and V2 determine T2 from the equation of state.
zenterix said:
For a reversible process we will have some ##T_{2,rev}## which may be different from ##T_2##. Thus, we aren't moving to the same final states necessarily.
Correct.
zenterix said:
(3) comes from writing out the differential form for ##U=U(T,V)## and assuming an ideal gas. Thus, for an ideal gas, (3) is "always" true. But is "always" really always here?
Yes, for a single phase of constant chemical composition.
zenterix said:
It seems not to matter that the irreversible process we have happens very suddenly and even violently. (3) expresses what happens when we change ##T## infinitesimally.
Yes, U is a function of T. The first law determines how U changes as a result of doing work and exchanging heat. The closure to the problem (i.e., the U side of the equation) is provided by the material behavior equation U = U(T) at thermodynamic equilibrium.
zenterix said:
Reversibility just seems to imply being able to sub in the ideal gas law for ##P_2## in (2) and (4).
In a reversible process, the system path consists of a continuous sequence of thermodynamic equilibrium states.
zenterix said:
I suppose any very rapid process can be considered as a sequence of infinitesimal steps in time, but we are not talking about time in these equations, we are talking about temperature. Temperature does not seem to be necessarily the same throughout the system in this rapid process.
There are also variations in pressure, and there are viscous stresses present in the gas which render the equation U=U(T) invalid for an irreversible path. The state of stress of the gas is not an isotropic pressure.
zenterix said:
4) With regard to equation (4), I have similar doubts. I guess volume can be seen, like time, to increase infinitesimally if we just break the process into small enough steps. And external pressure ##P_2## is constant, so (4) seems to make sense. But then we're back to my question 1) above.
Eqn. 4 is not valid for an irreversible path. The equation should read $$\Delta U=-\int{P_{ext}dV}$$assuming ##P_{ext}## has been specified explicitly as a function of V (e.g., ##P_{ext}=const## ).

You have asked excellent and thought provoking questions. Keep up the good work.

To get an idea of how the constitutive equation of a gas changes when we are considering an irreversible process (non-isotropic stress) vs a reversible process (isotropic stress P described by equation of state), Google "Newtonian fluid" , and see Chapter 1 of Transport Phenomena by Bird, Stewart, and Lightfoot. Newtonian fluids form the basis of virtually all Fluid Dynamics courses.

DeBangis21 and zenterix

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