The ΔS-sentials of Calculating Entropy Changes

DOFPro Team

The Second Law of Thermodynamics

• No apparatus can operate in such a way that its only effect (in system and surroundings) is to convert heat absorbed by a system completely into work.
• It is impossible by a cyclic process to convert the heat absorbed by a system completely into work.
• No process is possible which consists solely in the transfer of heat from one temperature level to a higher one.

Mathematical Statement of the Second Law

\(dW_{\mathrm{lost}} \geq 0\)

or

\(\Delta S_{\mathrm{system}} + \Delta S_{\mathrm{surroundings}} \geq 0\)

\(S\) can be thought of as the reversible normalized heat transfer or as the degree of randomness or disorder.

Remember:

\(\Delta S_{\mathrm{system}} \text{ or } \Delta S_{\mathrm{surroundings}}\)

can be less than zero. But

\(\Delta S_{\mathrm{system}} + \Delta S_{\mathrm{surroundings}} \geq 0\)

Calculation of ∆S for an ideal gas

Assume closed system of \(n\) moles initially at \(T_1\), \(P_1\) and finally at \(T_2\), \(P_2\)

First Law

\(dU = dQ + dW\)

\(d\hat{U} = \dfrac{dQ}{n} + \dfrac{dW}{n}\)

Now follow a reversible path (why?).

\(d\hat{U} = C_v dT\)

Calculation of ∆S for an ideal gas (cont.)

\(dS = \dfrac{dQ}{T}\ \) or \(\ Td\hat{S} = \dfrac{dQ}{n}\)

\(dW = -PdV\ \) or \(\ \dfrac{dW}{n} = - P d\hat{V}\)

Combining

\(C_vdT = T d\hat{S} - P d\hat{V}\)

\(d\hat{S} = \dfrac{C_v}{T} dT + \dfrac{P}{T} d\hat{V} = \dfrac{C_v}{T} dT + \dfrac{R}{\hat{V}} d\hat{V}\)

Calculation of ∆S for an ideal gas (cont.)

\(\Delta \hat{S} = \int\limits_{T_1}^{T_2} \dfrac{C_v}{T} dT + \int\limits_{\hat{V}_1}^{\hat{V}_1}\dfrac{R}{\hat{V}} d\hat{V} = \int\limits_{T_1}^{T_2} \dfrac{C_v}{T} dT + R \ln \dfrac{\hat{V}_2}{\hat{V}_1}\)

Alternatively

\(\ \ \ \ \Delta \hat{S} = \int\limits_{T_1}^{T_2} \dfrac{C_v}{T} dT + R \ln \dfrac{\hat{V}_2}{\hat{V}_1} = \int\limits_{T_1}^{T_2} \dfrac{C_p - R}{T} dT + R \ln \dfrac{\hat{V}_2}{\hat{V}_1}\)

\(\ \ \ \ \ \ \ \ \ = \int\limits_{T_1}^{T_2} \dfrac{C_p}{T} dT - \int\limits_{T_1}^{T_2} \dfrac{R}{T} dT + R \ln \dfrac{\hat{V}_2}{\hat{V}_1}\)

\(\ \ \ \ \ \ \ \ \ = \int\limits_{T_1}^{T_2} \dfrac{C_p}{T} dT - R \ln \dfrac{T_2\hat{V}_1}{T_1\hat{V}_2} = \int\limits_{T_1}^{T_2} \dfrac{C_p}{T} dT - R \ln \dfrac{P_2}{P_1}\)

Ideal Gas Example

Example: What is the minimum work required for an adiabatic compressor to compress \(10\ \mathrm{kg/s}\) of air from \(2\ \mathrm{kPa}\) and \(300\ \mathrm{K}\) to \(6\ \mathrm{MPa}\) (Assume ideal gas, \(C_p = (7 ⁄ 2)\mathrm{R}\) )?

First Law (neglect kinetic & potential)

\(\Delta \dot{H} = \dot{Q} + \dot{W}_s\)

\(\dot{m} \int_{T_1}^{T_2} C_p dT = \dot{W}_s = \dot{m} C_p \Delta T\)

Second Law

\(\Delta \dot{S}_\mathrm{surroundings} = 0\). Why?

IG Example (cont.)

\(\Delta \dot{S}_\mathrm{air} = \int\limits_{T_1}^{T_2} \dfrac{C_p}{T} dT - R \ln \dfrac{P_2}{P_1} = C_p \ln \dfrac{T_2}{T_1} - R \ln \dfrac{P_2}{P_1}\)

Minimum work for \(\Delta \hat{S} = 0\)

\(C_p \ln \dfrac{T_2}{T_1} = R \ln \dfrac{P_2}{P_1}\)

\(\dfrac{T_2}{T_1} = \left(\dfrac{P_2}{P_1}\right)^{\frac{R}{C_p}}\)

\(T_2 = T_1 \left(\dfrac{6000\ \mathrm{kPa}}{2\ \mathrm{kPa}}\right)^{2/7}\)

IG Example (cont.)

\(\dot{W}_s = \dot{m} C_p \Delta T\)

\(= \dfrac{1000\ \mathrm{g/s}}{29\ \mathrm{g/g-mol}}\left(\dfrac{7}{2}\right)(8.314\mathrm{ J/g-mol K})\left[\left(\dfrac{6000}{2}\right)^{2/7} - 1\right](300\ \mathrm{K})\)

\(=26.64\ \mathrm{MW}\)

Calculation of \(\Delta S\) for substances other than ideal gases

\(d\hat{S} = C_p \dfrac{dT}{T} - \dfrac{\partial \hat{V}}{\partial T} \bigg\vert_P dP\)

or

\(d\hat{S} = C_p \dfrac{dT}{T} - \beta \hat{V} dP\)

where the volume expansivity (like a thermal expansion coefficient) is

\(\beta \equiv \dfrac{1}{\hat{V}} \dfrac{\partial \hat{V}}{\partial T} \bigg\vert_P\).

It is usually small \(( \approx 0)\) for things other than gases.

Integrating C_p/T

Performing the integral:

\(C_p = a + b T +c T^2 + d T^3\)

\(\dfrac{C_p}{T} = \dfrac{a}{T} + b + c T + d T^2\)

\(\int\limits_{T_1}^{T_2} \dfrac{C_p}{T} dT = \int\limits_{T_1}^{T_2} \left(\dfrac{a}{T} + b + c T + d T^2\right) dT\)

\(= a \ln \dfrac{T_2}{T_1} + b ( T_2 - T_1) + \dfrac{c}{2} (T_2^2 - T_1^2) + \dfrac{d}{3} (T_2^3-T_1^3)\)

But what if \(C_p\) is in \(^\circ\mathrm{C}\), and the integral must be in \(\mathrm{K}\)? Review The Most Annoying Equation Conversion video and plug in

\(T(\mathrm{K})− 273.15 = T(\mathrm{K})− cf\)

for \(T\). Then

\(C_p = a' + b' T +c' T^2 + d' T^3\)

where \(T\) is in Kelvin and

\(a' = a - b \cdot cf + c \cdot cf^2 - d \cdot cf^3\)

\(b' = b - 2c \cdot cf + 3d \cdot cf^3\)

\(c' = c - 3d \cdot cf\)

\(d' = d\)

Summary

\(\Delta S\) for a heat reservoir:

\(\Delta S = \dfrac{Q}{T}\)

\(\Delta S\) for an ideal gas:

\(\Delta \hat{S} = \int\limits_{T_1}^{T_2} \dfrac{C_v}{T} dT + R \ln \dfrac{\hat{V}_2}{\hat{V}_1} = \int\limits_{T_1}^{T_2} \dfrac{C_p}{T} dT - R \ln \dfrac{P_2}{P_1}\)

\(\Delta S\) for a liquid or solid:

\(d\hat{S} = C_p \dfrac{dT}{T} - \beta \hat{V} dP \approx C_p \dfrac{dT}{T},\ \ \ \ \ \ \ \ \ \ \ \ \Delta \hat{S} \approx \int\limits_{T_1}^{T_2} \dfrac{C_p}{T} dT\)

\(\Delta S\) for water/steam: Use the steam tables.

The Takeaways

1. To calculate the change in entropy for a heat reservoir, calculate \(Q\) over \(T\).
2. To calculate the change in entropy for an ideal gas, either use the integral formula with \(C_v\) over \(T\) and \(\hat{V}\) or the integral formula with \(C_p\) over \(T\) and \(P\).
3. To calculate the approximate change in entropy for a liquid or a solid, integrate \(C_p\) over \(T\).
4. To calculate the change in entropy for steam or saturated liquid water, use the steam tables.

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