# Ideal Gas Processes

### Patrick McLaurin

• The First Law of Thermodynamics: Δ U = q + w     (1)

differential form: dU = dq + dw

• The change in internal energy, Δ U, of an ideal gas only depends on the change in temperature, Δ T. dU = n cV dT     (2)

Δ U = n cV Δ T as long as cV is constant over the relevant temperature range, Δ T

• Expansion work, w, is given by the change in volume, dV, pushing against the external pressure, Pext: dw = −Pext dV     (3)

Integration of this equation yields, w = −∫Pext dV.

## 1  Reversible Processes

Reversible processes are such that the system is always in equilibrium with the surroundings.

 Pext = Pgas ≡ P

This allows a substitution in equation (3) for work in terms of the gas pressure, P.

 dw = −Pext dV = −P dV     (4)

During a reversible process, the ideal gas law is always true for the gas, be it at initial state, final state, or along the path that connects those two states.

 PV = nRT

Using the subscript rev to remind us that the following equation is true for reversible processes only, dq = dqrev = T dS, which can be solved for the change in entropy, Δ S, by integration:

Δ S =
 dqrev T
(5)

### 1.1  Reversible & Isothermal Process

Isothermal processes are defined to have dT = 0 (Δ T = 0). This forces dU = Δ U = 0 by equation (2). Therefore, the magnitudes of q and w must be equal.

 dU = 0 = dq + dw  gives  dq = −dw  and therefore  q = −w

In order to calculate dq so that it can be integrated over T to give Δ S, the work, w, must be calculated first.
In the isothermal case, work accomplished by volume change ("expansion" work) will be accompanied by heat exchange where the exact amount of energy gained (lost) as heat is cancelled by energy lost (gained) in work, ensuring an isothermal process.
Equation (3) cannot be used to solve for work because the external pressure for a reversible, isothermal process cannot be constant. A constant Pext would have allowed pulling Pext out of the subsequent integral making it very easy to solve; oh well. What follows is an explanation of why the external pressure of a reversible, isothermal process cannot be constant.
Taking gas expansion as an example, in order for expansion to occur, the gas pressure must be higher than the external pressure, Pgas > Pext; this is accomplished reversibly by having infinitesimally small differences between the two. For the process to be isothermal, we can imagine that any heat the gas absorbs is immediately expended as work to expand against the external pressure, but in order for this to actually happen the external pressure must simultaneously decrease by just the right amount. Having expended the added heat as work towards expansion, the gas is now at a lower pressure due to the larger volume, and so the external pressure must match that lower gas pressure if the process is to remain reversible and isothermal.
Let us imagine what happens if the external pressure is set to be constant instead of decreasing during this expansion. To ensure isothermal behavior, you could suggest that any heat added to the gas could be converted to work to expand against the constant external pressure instead of causing a temperature increase in the gas. Such a process would allow the use of equation (3) to calculate work since the external pressure was defined to be constant and, therefore, we would presumably know its value at all points during the expansion. But, there’s an unphysical inconsistency; by increasing the volume of the gas, without increasing its temperature, we have decreased its pressure as required by the ideal gas law! This not only violates our reversibility, which requires Pgas = Pext, but it begs the question, how would this enlarged, lower pressure gas survive against the ever present force of the now higher, constant external pressure? It wouldn’t! In order to maintain pressure at the larger volume, Pgas = Pext, a gas expanding against constant external pressure requires more heat to be added than what gets expended as expansion work, so that the volume increase is offset by a temperature increase and the temperature increase would make the expansion NOT isothermal. In order for a gas expansion to be reversibly isothermal, as heat is absorbed by the gas the external pressure must decrease the correct amount so that the absorbed heat is immediately converted to work to expand against the decreasing external pressure.
Having proven that Pext must decrease to affect a reversible, isothermal expansion, we MUST solve for work by integrating both sides of equation (4) because while we don’t necessarily know what is happening to Pext in equation (3), we know what is happening to the gas and its pressure, P, in equation (4) because the gas must always obey the ideal gas law during a reversible process:

w = −
 V2 V1
P dV

The ideal gas law relates the pressure of the gas to its volume, P = nRT/V, which can be substituted into the equation above giving:

w = −
 V2 V1
 nRT V
dV

The moles of gas, n, universal gas constant, R, and temperature, T, are all constant for an isothermal process so they can be pulled out of the integral.

w = −nRT
 V2 V1
 1 V
dV

Solving this gives:

w = −nRT ln

 V2 V1

Recalling q = −w for isothermal ideal gas processes gives:

q = +nRT ln

 V2 V1

Finally, Δ S can be calculated now that we know how much heat was (reversibly) exchanged.

Δ S =
 qrev T
=
nRT ln

 V2 V1

T
= nR ln

 V2 V1

As an aside, the change in enthalpy, Δ H, is zero like Δ U because Δ T = 0.

 Δ H = Δ U + Δ (PV) = Δ U + Δ (nRT) = Δ U + nR Δ T = 0

Adiabatic processes are defined to have q = 0, therefore qrev = 0. Because qrev = 0, Δ S is also zero.

Δ S =
 qrev T
= 0

Like the isothermal case considered above, adiabatic expansion of a gas requires a reduction in the external pressure. Unlike isothermal expansion, adiabatic expansion results in a temperature decrease because no heat is absorbed to offset the energy lost through the expansion work.
Work cannot be calculated using equation (3) because Pext is changing during the process. Work cannot be calculated using equation (4) either because T is no longer a constant that can be pulled out of the integral after the ideal gas law substitution is made:

w = −nR
 V2 V1
 T V
dV

Calculating the work, w, requires solving for Δ U since they are equal, Δ U = q + w = 0 + w = w. Δ U is calculated from the change in temperature using equation (2).

 Δ U = n cV Δ T = w

Δ H can be obtained from the constant pressure heat capacity, cP = cV + R.

 Δ H = Δ U + Δ (PV) = Δ U + Δ (nRT) = n cV Δ T + nR Δ T = n (cV + R) Δ T = n cP Δ T

### 1.3  Reversible, Isochoric Process

Isochoric processes are defined to have Δ V = 0. Having no volume change leads to work being zero by equation (3).

 w = −Pext Δ V = 0

Calculating the heat, q, requires solving for Δ U since they are equal, Δ U = q + w = q + 0 = q. Δ U is calculated from the change in temperature using equation (2).
The entropy is calculated after making the substitution for dqrev = dq = dU = n cV dT.

Δ S =
 dqrev T
=
 T2 T1
 n cV dT T

Under the assumption that cV is constant over the range Δ T, the change in entropy is:

Δ S = n cV ln

 T2 T1

The change in enthalpy again relies on cp.

 Δ H = Δ U + Δ (PV) = Δ U + Δ (nRT) = n cV Δ T + nR Δ T = n (cV + R) Δ T = n cP Δ T

Since the volume isn’t allowed to change, this is an odd process to think about because it would require the external pressure to match the changes in gas pressure as heat is gained or lost without the gas interacting with the surroundings via a volume change.

### 1.4  Reversible, Isobaric Process

Isobaric processes are defined to have Δ P = 0. This requires Pgas = Pext = constant, which means any volume changes (work to be done) will result in temperature changes which manifest as exchange of heat. Therefore, the heat can be calculated from the constant pressure heat capacity, cP, and the change in temperature.

 qp = n cP Δ T

Δ S is obtained from substituting the expression for qp for qrev (in differential form) since the process is stated as being carried out reversibly.

Δ S =
 dqrev T
=
 T2 T1
 n cp dT T

Under the assumption that cp is constant over the range Δ T, the change in entropy is:

Δ S = n cP ln

 T2 T1

Having q, all that’s left to calculate the change in internal energy, Δ U, is work, w.

 Δ U = qP + w

Work can (finally!) be calculated using equation (3) because the external pressure is constant. Work can also be calculated from the difference between Δ U (obtained from equation (2)) and qP.

 w = Δ U − qP = n cV Δ T − n cP Δ T = −n R Δ T

This process occurs at constant pressure, so in this instance qP = Δ H.

 Δ H = Δ U + Δ (PV) = q + w + P Δ V + V Δ P =Δ P=0= qp − P Δ V + P Δ V + 0 = qP
 Δ H = Δ U + Δ (PV) = Δ U + Δ (nRT) = n cV Δ T + nR Δ T = n (cV + R) Δ T = n cP Δ T

This document was translated from LATEX by HEVEA.