The Carnot cycle provided a surprising conclusion: even a perfectly engineered heat engine cannot convert all heat into work. The maximum efficiency depends only on the temperatures of the hot and cold reservoirs.
Clausius wondered whether there might be a deeper thermodynamic quantity hiding within the analysis of the Carnot cycle. To investigate this possibility, let us examine the energy transfers that occur during the four reversible legs of the cycle.
Recall that a Carnot cycle consists of:
| Leg | Process | \(\Delta U\) | \(q\) | \(w\) |
|---|---|---|---|---|
| I | Isothermal expansion at \(T_h\) | 0 | \(nRT_h\ln\left( \frac{V_2}{V_1} \right)\) | \(-nRT_h\ln\left( \frac{V_2}{V_1} \right)\) |
| II | Adiabatic expansion | \(C_V(T_l-T_h)\) | 0 | \(C_V(T_l-T_h)\) |
| III | Isothermal compression at \(T_l\) | 0 | \(nRT_l\ln\left( \frac{V_4}{V_3} \right)\) | \(-nRT_l\ln\left( \frac{V_4}{V_3} \right)\) |
| IV | Adiabatic compression | \(C_V(T_h-T_l)\) | 0 | \(C_V(T_h-T_l)\) |
| Total | 0 | \(q_h + q_l\) | \(w_{\mathrm{tot}}\) |
The two adiabatic legs of the Carnot cycle connect the high-temperature isotherm to the low-temperature isotherm. Because the gas undergoes reversible adiabatic changes, the temperatures and volumes must satisfy
\[ TV^{R/C_V} = \text{constant} \]
Applying this relationship to the adiabatic expansion (Leg II) gives
\[ T_hV_2^{R/C_V} = T_lV_3^{R/C_V} \]
Applying it to the adiabatic compression (Leg IV) gives
\[ T_lV_4^{R/C_V} = T_hV_1^{R/C_V} \]
Rearranging both expressions:
\[ \left( \frac{V_3}{V_2} \right)^{R/C_V} = \frac{T_h}{T_l} \]
\[ \left( \frac{V_4}{V_1} \right)^{R/C_V} = \frac{T_h}{T_l} \]
Therefore,
\[ \boxed{ \frac{V_2}{V_1} = \frac{V_3}{V_4} } \]
This relationship is a direct consequence of the adiabatic legs and is the key mathematical result that allows the Carnot cycle to be simplified. Notice what happens when we make this substitution in the summary table. For illustrative purposes, we will also add a column for q/T.
| Leg | Process | \(\Delta U\) | \(q\) | \(w\) | \( \frac{q}{T} \) |
|---|---|---|---|---|---|
| I | Isothermal expansion at \(T_h\) | 0 | \(nRT_h\ln\left( \frac{V_2}{V_1} \right)\) | \(-nRT_h\ln\left( \frac{V_2}{V_1} \right)\) | \(nR \ln\left( \frac{V_2}{V_1} \right)\) |
| II | Adiabatic expansion | \(C_V(T_l-T_h)\) | 0 | \(C_V(T_l-T_h)\) | 0 |
| III | Isothermal compression at \(T_l\) | 0 | \(-nRT_l\ln\left( \frac{V_2}{V_1} \right)\) | \(nRT_l\ln\left( \frac{V_2}{V_1} \right)\) | \(-nR \ln\left( \frac{V_2}{V_1} \right)\) |
| IV | Adiabatic compression | \(C_V(T_h-T_l)\) | 0 | \(C_V(T_h-T_l)\) | 0 |
| Total | 0 | \(q_h + q_l\) | \(w_{\mathrm{tot}}\) | 0 |
Since the Carnot cycle is a closed cycle, the system returns to its initial state. Therefore,
\[ \Delta U_{\mathrm{tot}} = 0 \]
This is exactly what we expect because internal energy is a state function.
The total heat flow, however, is not zero (because \( T_h \) and \( T_l \) are different):
\[ q_h+q_l \neq 0 \]
Heat is clearly not a state function.
But now consider the final column of the table.
For the Carnot cycle, the efficiency derivation showed that
\[ \frac{q_h}{q_l} = -\frac{T_h}{T_l} \]
Rearranging gives
\[ \frac{q_h}{T_h} = -\frac{q_l}{T_l} \]
Therefore,
\[ \boxed{ \frac{q_h}{T_h} + \frac{q_l}{T_l} = 0 } \]
This is remarkable. The quantity \(q/T\) behaves exactly like a state function when summed around a reversible Carnot cycle.
Clausius recognized that this result was unlikely to be a coincidence.
If a quantity has a net change of zero around a reversible cycle, it may be the differential of a state function. He therefore proposed a new thermodynamic variable, called entropy, defined by
\[ dS = \frac{dq_{\mathrm{rev}}}{T} \]
The subscript "rev" is important because the argument relies on the reversibility of the Carnot cycle.
Big picture: The Carnot cycle revealed that while heat itself is not a state function, the quantity \(q/T\) behaves as though it were. This observation led Clausius to define a new thermodynamic state function: entropy.
Calculate the entropy change for \(1.00\ \mathrm{mol}\) of a monatomic ideal gas expanding isothermally from \(24.4\ \mathrm{L}\) to \(36.6\ \mathrm{L}\) at \(298\ \mathrm{K}\).
Entropy is defined by
\[ dS = \frac{dq_{\mathrm{rev}}}{T} \]
For a reversible isothermal expansion of an ideal gas,
\[ q_{\mathrm{rev}} = nRT\ln\left(\frac{V_2}{V_1}\right) \]
Therefore,
\[ \Delta S = \frac{q_{\mathrm{rev}}}{T} = nR\ln\left(\frac{V_2}{V_1}\right) \]
Substituting the values:
\[ \Delta S = (1.00\ \mathrm{mol}) \left( 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}} \right) \ln\left( \frac{36.6\ \mathrm{L}} {24.4\ \mathrm{L}} \right) \]
\[ \Delta S = 3.37\ \mathrm{J\,K^{-1}} \]
\[ \boxed{ \Delta S = +3.37\ \mathrm{J\,K^{-1}} } \]
Physical interpretation: The entropy change is positive because the gas expands into a larger volume. The same amount of energy is now spread over a larger region of space, giving the molecules more possible positions and more ways to arrange themselves.
This agrees with the qualitative idea that entropy measures the dispersal of energy and is related to the randomness or chaotic nature of the system. When a gas expands, the molecules have more accessible space, so an increase in entropy makes sense.
During compression, the opposite occurs. The molecules are confined to a smaller volume, the number of accessible positions decreases, and the entropy change for the gas is negative.
Calculate the entropy change for \(1.00\ \mathrm{mol}\) of a monatomic ideal gas \(\left(C_V=\frac{3}{2}R\right)\) that is heated from \(298\ \mathrm{K}\) to \(447\ \mathrm{K}\) while occupying a constant volume of \(36.6\ \mathrm{L}\).
Entropy is defined by
\[ dS = \frac{dq_{\mathrm{rev}}}{T} \]
For a constant-volume process,
\[ dq_{\mathrm{rev}} = nC_V\,dT \]
and therefore
\[ dS = \frac{nC_V\,dT}{T} \]
Integrating from the initial temperature to the final temperature:
\[ \Delta S = \int_{T_1}^{T_2} \frac{nC_V}{T}\,dT \]
Since \(C_V\) is constant,
\[ \Delta S = nC_V \int_{T_1}^{T_2} \frac{dT}{T} \]
\[ \Delta S = nC_V \ln\left( \frac{T_2}{T_1} \right) \]
For a monatomic ideal gas,
\[ C_V = \frac{3}{2}R = \frac{3}{2}(8.314) = 12.47\ \mathrm{J\,mol^{-1}\,K^{-1}} \]
Substituting the values:
\[ \Delta S = (1.00\ \mathrm{mol}) (12.47\ \mathrm{J\,mol^{-1}\,K^{-1}}) \ln\left( \frac{447\ K}{298\ K} \right) \]
\[ \Delta S = 5.06\ \mathrm{J\,K^{-1}} \]
\[ \boxed{ \Delta S = +5.06\ \mathrm{J\,K^{-1}} } \]
Test of reasonableness: The gas is heated while its volume remains constant. Heating increases the kinetic energy of the molecules and spreads the system's energy over a larger number of accessible molecular states. Because the energy becomes more dispersed, we expect the entropy to increase. The positive value of \(\Delta S\) is therefore exactly what we would anticipate.
If the gas had instead been cooled from \(447\ \mathrm{K}\) to \(298\ \mathrm{K}\), the logarithm would be negative and the entropy change would likewise be negative, reflecting the reduced dispersal of energy.
Calculate the entropy change for \(1.00\ \mathrm{mol}\) of a monatomic ideal gas \(\left(C_p=\frac{5}{2}R\right)\) that expands at constant pressure from \(24.4\ \mathrm{L}\) and \(298\ \mathrm{K}\) to \(36.6\ \mathrm{L}\) and \(447\ \mathrm{K}\).
For a constant-pressure process,
\[ dq_{\mathrm{rev}} = nC_p\,dT \]
Therefore,
\[ dS = \frac{dq_{\mathrm{rev}}}{T} = \frac{nC_p\,dT}{T} \]
Integrating from \(T_1\) to \(T_2\):
\[ \Delta S = \int_{T_1}^{T_2} \frac{nC_p}{T}\,dT \]
Since \(C_p\) is constant,
\[ \Delta S = nC_p \ln\left(\frac{T_2}{T_1}\right) \]
For a monatomic ideal gas,
\[ C_p = \frac{5}{2}R = \frac{5}{2}(8.314) = 20.79\ \mathrm{J\,mol^{-1}\,K^{-1}} \]
Substituting the values:
\[ \Delta S = (1.00\ \mathrm{mol}) (20.79\ \mathrm{J\,mol^{-1}\,K^{-1}}) \ln\left(\frac{447}{298}\right) \]
\[ \Delta S = 8.43\ \mathrm{J\,K^{-1}} \]
\[ \boxed{\Delta S = +8.43\ \mathrm{J\,K^{-1}}} \]
Test of reasonableness: The gas is heated and expands to a larger volume. Both effects increase entropy: the energy is more highly dispersed at the higher temperature, and the molecules have more accessible space. Therefore, a positive entropy change is expected.
Notice that this value is larger than the constant-volume heating example. At constant pressure, the gas expands while it is heated, so the entropy increase includes both the effect of increasing temperature and the effect of increasing volume.
Calculate the entropy change for \(1.00\ \mathrm{mol}\) of a monatomic ideal gas \(\left(C_V=\frac{3}{2}R\right)\) that undergoes a reversible adiabatic expansion from \(24.4\ \mathrm{L}\) at \(298\ \mathrm{K}\) to \(36.6\ \mathrm{L}\).
Entropy is defined by
\[ dS = \frac{dq_{\mathrm{rev}}}{T} \]
For an adiabatic process,
\[ dq = 0 \]
and therefore
\[ dS = \frac{0}{T} = 0 \]
Integrating from the initial state to the final state:
\[ \Delta S = \int dS = \int 0 \]
\[ \boxed{ \Delta S = 0 } \]
Test of reasonableness: At first glance, this result may seem surprising. The gas expands to a larger volume, which by itself would tend to increase entropy. However, because the expansion is adiabatic, the gas also cools as it expands. The decrease in entropy associated with the lower temperature exactly offsets the increase in entropy associated with the larger volume.
As a result, a reversible adiabatic process is also an isentropic process:
\[ \boxed{ \Delta S = 0 } \]
This fact will be extremely useful when analyzing heat engines and refrigerators because the adiabatic legs of a Carnot cycle contribute no entropy change.
A pathway will be chosen at random. Use the given data to calculate \(\Delta S\) for \(1.00\ \mathrm{mol}\) of a monatomic ideal gas.
Big picture: The Carnot cycle revealed that the quantity \(q/T\) behaves like a state function for reversible processes, leading Clausius to define entropy. Entropy provides a quantitative measure of energy dispersal and will ultimately become the key quantity used to determine whether a process is spontaneous.