Thermodynamics: External vs. Internal, Real vs. Ideal

Has anyone ever had any doubt about the relation between internal and external work in regards to thermodynamics?

In consideration of irreversible processes, why is the work done measured by the work done against the external pressure and not the work done by the internal pressure?

If the forces or pressures are not balanced the system is not in thermodynamic equilibrium while the process is ongoing. If the internal pressure is greater than the external force you have a dynamic rather than a quasi-static process. This means that some of the energy of the system is apparently kinetic. Applying conservation of energy (first “law” of thermodynamics), you have to take into account the kinetic energy as part of the work done. This way, the first “law” of thermodynamics is maintained.  For example, in the case that an explosion of a gas in a cylinder doing work against a piston, the internal pressure produces kinetic energy of the piston and of the molecules of the exploding gas. So, suppose the piston gets lifted by the expansion from the exploding gas. You’ll arrive at a point that the internal and external pressure become equal (by the assumption that the cylinder has a substantial length), but because the piston is moving with some speed it’ll go higher before it stops. At that point, the external pressure is higher than the internal pressure and the piston will come back down, picking up the speed. It goes past the point where internal and external pressures are equal and eventually will come to a stop when its kinetic energy is used up in compressing the gas in the cylinder. It’ll repeat the oscillation until all of the energy is used up in the frictional form of heat.

Applying the first “law” of thermodynamics now to the stationary system using the external pressure to determine the work done, Work = PdV, and the final temperature (with the extra heat added by the conversion of kinetic energy into heat). In this way, the first “law” of thermodynamics is maintained.

P_{int}\Delta V = P_{ext}\Delta V + KE

Up above, you’ll see that the internal pressure does indeed work against the external pressure and it does additional work in adding kinetic energy to the system. After everything has returned to equilibrium and the kinetic energy has dissipated (the process of conversion) into heat in the system, ideally, this is when you’d want to analyze the system. The result is the same from the perspective of energy as if it had occurred under a quasi-static process. Now, if you decide to analyze the system before things have returned to equilibrium, you’ll have to use dW = P_{int}dV for the work that’s been done. This results in work done against the external pressure plus kinetic energy in the dynamic system. dV is based on the internal pressure. Thermodynamics divides the universe into two distinctive parts: the system and the surroundings (of that system). Both the system and its surroundings are, themselves, divided by a boundary. A system and its surroundings have state variables that refer exclusively to either the system or the surroundings. Changes in state variables are wholly dependent upon both the start and end points and not the intermediate path. These variables account for the exchanges across the boundary. None of these are state variables since they do not describe the state of anything. A prime assumption of thermodynamics is that if a quantity is defined by one of these exchange variables is passed across the boundary it is the same on both sides of the boundary.

Work done by a system’s surroundings is equal to the work received by the system. The reverse is the same.

W_system = W_surroundings = W

As presented, there are two opportunities to calculate the value of exchange variables, either from the system or from the system’s surroundings, because they refer to the same exact quantity. For the system, the work, w = {\int_pdV} if the system pressure can be defined at all times during the process. For the system’s surroundings, w = {\int_pdV}, again, but since the pressure is constant w = p\int_dv = p(v_2 - v_1). Of the state variables one can claim that the volume change in the system is equivalent to the volume change in the system’s surroundings.

Thermodynamics is, essentially, a study of changes occurring between equilibrium states. Gas can do work on two things:  the system and the system’s surroundings. As it undergoes a quasi-static thermodynamic process all of the work is then done on the system’s surroundings. If the process is dynamic [read: irreversible] it’ll do some work on the system so it does not do all of its work on the surroundings. When equilibrium is reached, the work done on the system (which includes the gas itself) ends up as thermal energy in the system. Thermal energy is converted into mechanical work, some of which returns to the system as thermal energy. If we are to look at the system in its initial and final equilibrium states, we’ll notice that the work done on the system that ends up back as thermal energy, in the system, is a wash, or treated the same as if it had remained as heat. You see, the work done that affects the state of the system [and the system’s surroundings] is the work done against an external pressure.

Just think of the adiabatic free expansion of an ideal gas through a nozzle into an empty chamber. There’s no external work done [on the system]. The process is dynamic. Internally, the gas pressure is unbalanced and causes a stream of gas molecules to exit the nozzle at a high velocity. Once gas has entered the empty chamber it’ll provide a certain pressure that the incoming steam will have to do work against. So, the gas does work–work on itself. As the gas eventually reaches equilibrium, the work that it has done on itself ends up as thermal energy in the gas and its final temperature is the same as its initial temperature. It’ll appear as if no work had been done at all by the gas.

To speak a little more on the pressure, let’s say that initially the gas pressure is at P and the external pressure is a constant P(ext). The pressure P > P(ext) so that the gas will expand until its pressure P = P(ext). You’ll see that the work done by the gas–in infinitely small intervals–will be the PdV, which is a positive quantity, however, this pressure is undefined since the process is not quasi-static, so the PdV is intrinsically impossible to calculate. Yet, on the other hand, the work done by the system’s surroundings will equal the negative of the work done by the system but having a defined value -P_{ext}dV, this can be calculated.

|W_{sys} | = |PdV| = |W_{sur}| = |-P_{ext}dV| so that || P || = || P_{ext}||

What can be said about the calculation above is that the pressure is not undefined since any individual part does not have a pressure, it is undefined because every individual part (in the case of infinitesimal volume) has a differing pressure for which you’ll need to take an average of the pressures. If the pressures were identical there would be no expansion of the gas. If the internal pressure is higher than that of the external pressure, the system would not be in equilibrium. This is why the system expands, changing both the volume and the pressure simultaneously. In regards to the system’s surroundings. Let’s take the environment to be so large that its volume contraction is relatively tiny and it will not alter the pressure of the system’s surroundings, therefore, the pressure of the system’s surroundings is constant.

In the adiabatic sense of thermodynamics, if you have the mass and the specific heat and the energy, you can use…

\Delta E = mc\Delta T

…to determine the change of temperature of the system and the system’s surroundings. Adiabatic processes (the processes that are reversible) are those in which no heat is exchanged (entropy), thus the change in internal energy is given by…

dU = - p \,dV

…from the first “law” of thermodynamics. Neither the pressure, temperature or volume is constant in an adiabatic process but the combination…

p V^\gamma This particular condition, with the ideal gas law, will permit an integration of the above expression. The internal energy is a state function (of the previously mentioned state variables that comprise the system) and for an ideal gas, that is strictly dependent only on temperature. Irrespective of all other considerations, the change in internal energy for any process that starts at…


…and ending with…


…can simply be given by…

\Delta U = \int^{T_f}_{T_i} C_V \,dT

Constant volume, C_V, is a state function and once it’s shown that two state functions are equal (it doesn’t matter what sort of process is utilized in order to show it), the equality is valid in general. Heat is not a state function, therefore, partial derivatives of state variables cannot be taken from the exchange of heat. For example, according to the first “law” of thermodynamics, heat capacity, at a constant volume, is defined as…

\frac{\partial U} {\partial T}

…and that’s all. What’s irrelevant is the fact that heat capacity, at a constant volume, is not related to the heat exchanged in an adiabatic process, however, it’s always the partial derivative of the internal energy in respect to temperature. Understand that the internal energy is dependent upon the temperature alone, though the heat capacity, at a constant volume, to calculate the change in internal energy. For constant pressure, C_P, the ideal gas law clearly states that…

p \frac{\partial V} {\partial T} = nR

The result is even more direct when considering the first “law” of thermodynamics for a constant volume process:

dU = \delta Q

Definition-wise, \delta Q = C_V\, dT is at constant volume, so clearly C_V = \frac{\partial U} {\partial T}. That is the definition of constant volume, C_V. There isn’t all that much more to it than this. The internal energy, nR, is a state function and so are its corresponding partial derivatives with respect to temperature. Regard this as a proof of the statement that…

C_P = C_V + nR

…for constant pressure processes since constant pressure, C_P and constant volume, C_V are both state functions.

In reference to how reactive something can be, let’s take sulfuric acid as an example. Sulfuric acid reacts chemically with water, H_2O, which is the source of heat. Presumably, we could be talking about ink. We can stipulate that ink does not react and that the enthalpy that mixing for the ink-water system is zero. Referring back to gas, consider the flow of an ideal gas through a horizontal pipe. The pressure drops over some length of the flow through the horizontal pipe–and we’ll just assume that the pipe is insulated–and there is no work done on or by the gas on the environment. How does the gas change state as it flows? You’ll have to apply the first “law” of thermodynamics and take into consideration of what happens afterwards. Perhaps, you can draw a [stationary] control volume around the pipe, especially if there’s a mass flow in and flow out that one can deem as the easiest way. Or, you can draw a control volume around a given mass that travels down the pipe (i.e., control mass). The situation I’m describing is known as the Joule-Thompson Coefficient/Effect.

When a real gas, differentiated from an ideal gas, expands at constant enthalpy. In other words, no heat is transferred to or from the gas, and no external work is extracted. The gas will be either cooled or heated by the expansion. That change in gas temperature in conjunction with the change in pressure is called the Joule-Thompson Coefficient and it’s denoted by \mu, in definition:

\mu = dT/dP at constant enthalpy

The value of µ depends on the specific gas as well as the temperature and pressure of the gas before expansion. For all real gases, \mu will equal zero at some point which is called the “inversion point”. If the gas temperature is below its inversion point temperature, \mu is positive, and if the gas temperature is above its inversion point temperature, \mu is negative. d_P is also negative when a [real] gas expands.


If the gas temperature is below its inversion temperature:

  • \mu is positive and d_P is always negative
  • Hence, the gas cools since d_T must be negative

If the gas temperature is above its inversion temperature:

  • \mu is negative and d_P is always negative
  • Hence, the gas heats since d_T must be positive

Perry’s Chemical Engineers’ Handbook provides tabulations of \mu versus temperature and pressure for several gases as do many other reference books. For most gases, at atmospheric pressure, the inversion temperature is fairly high, at least above room temperature. Most gases at those temperature and pressure conditions are cooled by isenthalpic expansion.

For instance, helium and hydrogen are two gases whose Joule-Thompson inversion temperatures at atmospheric pressure are considerably low, I’m talking around -222°C (for helium). Thus, helium and hydrogen will warm when they’re expanded at constant enthalpy at atmospheric pressure and average room temperatures.

The Joule-Thompson inversion temperature is not only a function of pressure but also of temperature. The inversion temperature is a function of the physical state.

One thing that stands out is that the work the gas does is not going anywhere. It’s probably going back into the gas. Consider a well-insulated cylinder fitted with a piston that’s placed inside a pipe at a location where the pressure is 100 psig (pounds per square inch gauge). Rolling with the assumption that air is in the cylinder and also in the pipe. The cylinder will be allowed to travel along the length of the pipe to where the pipe discharges to atmosphere at 0 psig. The work done by the air in the cylinder must go into the flow stream. If the cylinder is well-insulated (adiabatic) and the expansion is isentropic as we would expect it to be, the air in the cylinder cools significantly. In comparison, the air in the pipe won’t cool significantly at all. Then, it would be pretty close to being classified as an ideal gas and the change in internal energy for the air traveling down the pipe is essentially zero. Therefore, there would be no change unless there’s heat/work added to or removed from the stream of flow.

In the first case, we have a traveling cylinder that does work P_dV and in the second case, we have air that does no work on the system’s surroundings yet undergoes an identical change in pressure with no change in its internal energy. So, does the air flowing through the pipe do work on itself? If so, it does work exactly equal to the work needed to keep the internal energy change at zero. At lease, that’s what can be assumed.

Simply speaking, it’s a conservation of energy.




11 thoughts on “Thermodynamics: External vs. Internal, Real vs. Ideal

  1. this was a really quality post. i wasn’t aware of the many ripples and depth to this story until i surfed here through google! great job.


  2. It is really a nice and useful piece of info. I am happy that you shared this useful information with us. Please keep us informed like this. Thank you for sharing.


  3. I do not even know how I ended up here, but I thought this post was great.
    I do not know who you are but definitely you are going to a famous blogger if you are
    not already 😉 Cheers!


  4. Hi, I do believe this is a great web site. I stumbledupon
    it 😉 I will return once again since I book marked it.
    Money and freedom is the best way to change, may you be rich
    and continue to guide other people.


Leave a Reply

Please log in using one of these methods to post your comment: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.