The First Law: Expansion and Engines

Shaun Williams, PhD

Expansion of Gases

Isothermal Expansion

• Consider the work done in the reversible expansion between two thermodynamic states
  • \(V_2 \gt V_1\)
• The first law states $$ dE = \dbar q + \dbar w =T\,dS - P\,dV+\mu\,dn $$
• If we restrict ourselves to a closed system then \(dn = 0\) so $$ dE=\dbar q+\dbar w = T\,dS-P\,dV $$

Our Expansion Setup

More Simplifications

• If our process is reversible then from the last chapter \(\dbar q_{rev} = T\,dS\)
• This means that \(\dbar w_{rev}=-P\,dV\)
• Let’s say this is an isothermal expansion
  • Using a heat reservoir and diathermal walls $$ w_{T,rev} = \int_{V_1}^{V_2} \dbar w_{T,rev} = -\int_{V_1}^{V_2} P\,dV $$

Reversible vs Irreversible Expansions

Applying the Ideal Gas Law

• From the ideal gas law we know that \(P=\frac{nRT}{V}\)
• Using this we find that $$ \begin{align} w_{T,rev} &= -\int_{V_1}^{V_2} P\,dV = -nRT \int_{V_1}^{V_2}\frac{dV}{V} \\ &= -nRT \ln \left(\frac{V_2}{V_1}\right) \end{align} $$

Example 8.1

A vacuum chamber in a spectrometer is maintained at an operating pressure of \(10.0\,\mathrm{mtorr}\) by a two-stage rotary vane pump with an exhaust pressure at the pump outlet of \(800.\,\mathrm{torr}\). What is the minimum power in watts (\(\mathrm{\bfrac{J}{s}}\)) consumed by the pump to keep the chamber at this pressure when there is a flow of \(0.22\,\mathrm{\bfrac{mmol}{s}}\) and \(T=300\,\mathrm{K}\)?

For an Irreversible Case

• If we have an irreversible expansion with a constant external pressure of \(P_{min}\) $$ \begin{align} w_{T,irr} &= \int_{V_1}^{V_2}\dbar w_{irr} \\ &= -\int_{V_1}^{V_2} P_{min}\,dV \\ &= -P_{min} \left(V_2-V_1\right) \end{align} $$

Plots of Pressure vs Volume for Reversible vs Irreversible Expansions

The Balance of Energy Flow in Isothermal Expansion

More Ideal Gas Approximations

• For an ideal gas we know that \(E=\frac{3}{2}nRT\)
• This means that for a closed, isothermal expansion, \(\Delta E=0\)
• Therefore $$ \begin{align} \Delta E = q+w &=0 \\ q_{T,rev} - nRT\ln \left(\frac{V_2}{v_1}\right) &= 0 \\ q_{T,rev} &= nRT \ln \left( \frac{V_2}{V_1}\right) \\ q_{T,irr} &= P_{min} \left(V_2-V_1\right) \end{align} $$

Something to Notice

• Even though this is a reversible process and \(dS_T=0\) $$ \dbar q_{rev} \ne 0 $$
• This means that for the system $$ dS = \frac{\dbar q}{T} \ne q $$
• This means that the reservoir must respond with its own entropy change that exactly cancels the sample’s entropy change

Reversible Adiabatic Expansion

• In an adiabatic expansion, no heat can flow through the boundary walls, thus $$ \dbar q_{rev} = T\,dS =0 $$
• This means that \(dE=\dbar w_{S,rev} = -P\,dV\)
• Continuing to assume only \(PV\) work and that we have a closed system $$ dE = -P\,dV = -\frac{nRT\,dV}{V} $$

Problems!

$$ dE=-P\,dV = -\frac{nRT\,dV}{V} $$

• Because the sample container is adiabatic, the sample temperature can change during the expansion
• If the temperature changes, which it will, the final volume will also change
• We need another equation so we can tell how the temperature and volume change

Welcome Back Heat Capacity!

• The heat capacity relates \(\Delta E\) and \(\Delta T\)
• For an ideal gas \(E=\frac{1}{2}N_{ep}nRT\)
• Now let’s do some math $$ \begin{align} dE &= \left(\frac{\partial E}{\partial T}\right)_VdT=C_V\,dT \\ C_V\,dT &= -\frac{nRT\,dV}{V} \\ \int_{T_1}^{T_2} \frac{C_V\,dT}{T} &= -\int_{V_1}^{V_2} \frac{nR\,dV}{V} \\ C_V\ln\frac{T_2}{T_1} &= -nR\ln \frac{V_2}{V_1} \\ V_2 &= V_1 \left(\frac{P_2}{P_1}\right)^{-\frac{C_V}{C_P}} \end{align} $$

More Expansion Stuff

• The work done during the expansion is equal to $$ \Delta E = C_V \Delta T $$
• Thus $$ \begin{align} w_{S,rev} &= C_V (T_2 - T_1) \\ &= C_V \left[ \frac{P_2V_2}{nR}-\frac{P_1V_1}{nR}\right] \\ &= \frac{C_V}{nR}\left[P_2V_1\left(\frac{P_2}{P_1}\right)^{-\bfrac{C_V}{C_P}}-P_1V_1\right] \\ &= \frac{C_V}{nR}(P_1V_1)\left[\frac{P_2}{P_1}\left(\frac{P_2}{P_1}\right)^{-\bfrac{C_V}{C_P}}-1\right] \end{align} $$

More Derivation

• Continuing on $$ \begin{align} w_{S,rev} &= \frac{C_V}{C_P-C_V}(P_1V_1)\left[\left(\frac{P_2}{P_1}\right)^{1-\bfrac{C_V}{C_P}}-1\right] \\ &= \frac{1}{\bfrac{C_P}{C_V}-1} (P_1V_1)\left[\left(\frac{P_2}{P_1}\right)^{1-\bfrac{C_V}{C_P}}-1\right] \\ &= \frac{P_1V_1}{\gamma-1}\left[\left(\frac{P_2}{P_1}\right)^{1-\bfrac{1}{\gamma}}-1\right] \end{align} $$
• Note the heat capacity ratio \(\gamma = \bfrac{C_P}{C_V}\)

Three Plots

Compression Ratios and Work

Example 8.2

If \(1.0\,\mathrm{L}\) of hexamethylene triperoxide diamine (HMTD) explodes by suddenly decomposing to gases at a pressure of \(1.0\,\mathrm{kbar}\) at the ambient temperature, calculate the grams of TNT that would release a pressure wave of equal energy, assuming \(4680\,\mathrm{\bfrac{J}{g\,TNT}}\), and assuming the explosion is reversible. Assume an average value of \(\gamma\) of \(1.4\), an ambient pressure of \(1.0\,\mathrm{bar}\), and that 70% of the energy is in the pressure wave.

Special Case

• Let’s look at an adiabatic expansion in which the enthalpy is held constant while \(P\), \(V\), and \(T\) can change
• A gas is in one chamber separated by a permeable plug
• A piston compresses the gas in the chamber forcing it through the plug

The Math

• No heat can flow so \(\Delta E=w_2+w_1\)
• Because we kept the pressure in each container constant the work in each container is just \(–P\Delta V\) $$ \Delta E = -P_2V_2 - P_1(-V_1) = P_1V_1-P_2V_2 $$
• Look at the enthalpy $$ \begin{align} \Delta H &= \Delta (E+PV) \\ &= \Delta E + \Delta (PV) \\ &= (P_1V_1-P_2V_2) + (P_2V_2-P_1V_1) \\ &= 0 \end{align} $$

The Special Expansion

• This special type of adiabatic expansion is called the Joule-Thomson expansion
• Named for James Joule and William Thomson
  • Both of whom have very common units named after them

Temperature Change?

• What happens to the temperature of the gas as it passes through the plug between pressures
• We are looking for the Joule-Thomson coefficient \(\left(\frac{\partial T}{\partial P}\right)_H\) $$ \left(\frac{\partial T}{\partial P}\right)_H=-\left(\frac{\partial T}{\partial H}\right)_P\left(\frac{\partial H}{\partial P}\right)_T = -\frac{1}{C_P}\left(\frac{\partial H}{\partial P}\right)_T $$
• Simplifying $$ \begin{align} \left(\frac{\partial H}{\partial P}\right)_T &= \left( \frac{T\partial S+V\partial P}{dP}\right)_T = \left( \frac{T\partial S}{dP}\right)_T+\left(\frac{V\partial P}{dP}\right)_T \\ &= T\left(\frac{\partial S}{\partial P}\right)_T+V \\ &= -T\left(\frac{\partial V}{\partial T}\right)_P+V=-TV\alpha +V \\ \left(\frac{\partial T}{\partial P}\right)_H &= \frac{TV\alpha -V}{C_P} \end{align} $$

Joule-Thomson Coefficient for a Real Gas

• Consider a gas governed by the second virial expansion \(P=RT\left[V_m^{-1}+B_2(T)V_m^{-2}\right]\) $$ V=\frac{nRT}{P}\left(1+\frac{n}{V}B_2(T)\right) $$
• thus $$ \begin{align} V\alpha &= \left(\frac{\partial V}{\partial T}\right)_P \\ &= \frac{nR}{P}\left(1+\frac{n}{V}B_2(T)\right) +\frac{nRT}{P}\left[\begin{split} & -\frac{n}{V^2}B_2(T)\left(\frac{\partial V}{\partial T}\right)_P \\ &+\frac{n}{V}\left(\frac{\partial B_2}{\partial T}\right)_P\end{split}\right] \\ &= \frac{V}{T}+\frac{nRT}{P}\left[-\frac{n}{V^2}B_2(T)\left(\frac{\partial V}{\partial T}\right)_P+\frac{n}{V}\left(\frac{\partial B_2}{\partial T}\right)_P\right] \end{align} $$

Imagine A LOT of Math

• After a lot more algebra, calculus, etc we arrive at $$ \left(\frac{\partial T}{\partial P}\right)_H = \frac{T\left(\frac{\partial B_2}{\partial T}\right)_P-B_2(T)}{C_{Pm}} $$
• In terms of the van der Waals coefficients $$ \left(\frac{\partial T}{\partial P}\right)_H = \frac{\frac{2a}{RT}-b}{C_{Pm}} $$

Limits of the Equation

• At high temperatures $$ \left(\frac{\partial T}{\partial P}\right)_H \approx -\frac{b}{C_{Pm}} $$
• At low temperatures $$ \left(\frac{\partial T}{\partial P}\right)_H \approx \frac{2a}{RTC_{Pm}} $$

Joule-Thomson Experiment Different Limits

1. Attractive forces dominate
2. Attractive forces are insignificant

Interesting Expansion Effects

• Methane in compressed tanks cool as it expands through the nozzle
• \(\chem{H_2}\) and \(\chem{He}\), however, have very weak intermolecular forces – \(a\) is very small
  • The Joule-Thomson coefficient is negative
  • The gases warm as is expands
• The temperature at which the Joule-Thomson coefficient changes sign is called the Joule-Thomson inversion temperature

Example 8.3

Use the van der Waals constants to estimate the Joule-Thomson coefficients for \(\chem{He}\), \(\chem{H_2}\), \(\chem{N_2}\), and \(\chem{CO_2}\) at \(300.\,\mathrm{K}\).

Example 8.4

Given \(a=0.034\,\mathrm{L^2\,bar\,mol^{-2}}\) and \(b=0.0237\,\mathrm{L\,mol^{-1}}\) for \(\chem{He}\), and \(2.25\,\mathrm{L^2\,bar\,mol^{-2}}\) and \(0.0428\,\mathrm{L\,mol^{-1}}\) (respectively) for \(\chem{CH_4}\), calculate the Joule-Thomson inversion temperatures for these gases.

Engines

A Heat Engine

• A heat engine is a device that continuously converts heat into work
• Consider a machine that cycles among \(A \rightarrow B \rightarrow C \rightarrow D \rightarrow A\) $$ w=-\int_A^B P\,dV - \int_B^C P\,dV - \int_C^D P\,dV - \int_D^A P\,dV $$

The Carnot Cycle

• One of the simplest cycles is the Carnot cycle
• Named for Nicolas Carnot
  • Developed this first model of the heat engine in 1824

The First of Four Steps of the Carnot Cycle

• Isothermal expansion
• The gas expands isothermally at temperature \(T_{hot}\) from pressure \(P_A\) and volume \(V_A\) to \(P_B\) and \(V_B\)

The Second of Four Steps of the Carnot Cycle

• Adiabatic expansion
• Expel the steam
• The system cools
• Expands adiabatically until it reaches \(T_{cold}\) at \(P_C\) and \(V_C\)

The Third of Four Steps of the Carnot Cycle

• Isothermal compression
• The gas is compressed isothermally at temperature \(T_{cold}\) to \(P_D\) and \(V_D\)

The Final of Four Steps of the Carnot Cycle

• Adiabatic compression
• Purge the cold water
• Return to the starting conditions

Step 1: Isothermal Expansion

• First we can work out the work in this step $$ \begin{align} w_1 &= -\int_{V_A}^{V_B} P(V)\,dV = -\int_{V_A}^{V_B} \frac{nRT_{hot}}{V}\,dV \\ &= -nRT_{hot} \ln \frac{V_B}{V_A} \end{align} $$
• Since this is isothermal (and ideal gas) $$ \begin{align} \Delta E(T) &= q+w =0 \\ q_1 &= -w_1 = nRT_{hot} \ln \frac{V_B}{V_A} \end{align} $$

Step 2: Adiabatic Expansion

• For an ideal gas \(\Delta E = C_V\Delta T\)
• For adiabatic process, \(q=0\), so \(\Delta E=w\) $$ \begin{align} w_2 &= C_V(T_{cold}-T_{hot}) \\ q_2 &= 0 \end{align} $$

Step 3: Isothermal Compression

• Analogous to step one, only compression now $$ \begin{align} w_3 &= -\int_{V_C}{V_D} P(V)dV=-nRT_{cold}\ln\frac{V_D}{V_C} \\ q_3 &= nRT_{cold}\ln \frac{V_D}{V_C} \end{align} $$

Step 4: Adiabatic Compression

• Analogous to step 2 $$ \begin{align} w_4 &= -C_V(T_{cold}-T_{hot}) \\ q_4 &= 0 \end{align} $$

Total Work Done

• We can add up all the work done $$ \begin{align} w &= w_1+w_2+w_3+w_4 \\ &= -nR\left(T_{hot}\ln \frac{V_B}{V_A}+T_{cold}\ln \frac{V_D}{V_C}\right) \end{align} $$
• We also know that \(C_V\ln \frac{T_2}{T_1}=-nR\ln\frac{V_2}{V_1}\)
• This means that $$ C_V\ln\frac{T_{hot}}{T_{cold}}=-nR\ln\frac{V_B}{V_C} = -nR\ln\frac{V_A}{V_D} $$

Further Simplification

• We can relate the volume ratios $$ \frac{V_D}{V_A} = \frac{V_C}{V_B} \text{ and } \frac{V_D}{V_C}=\frac{V_A}{V_B} $$
• So the total work is now $$ \begin{align} w &= -nR \left( T_{hot} \ln \frac{V_B}{V_A} + T_{cold} \ln \frac{V_D}{V_C} \right) \\ &= -nR \left( T_{hot} \ln \frac{V_B}{V_A} + T_{cold} \ln \frac{V_A}{V_B} \right) \\ &= -nR \left( T_{hot} - T_{cold} \right) \ln \frac{V_B}{V_A} \\ q &= nR \left( T_{hot} - T_{cold} \right) \ln \frac{V_B}{V_A} \end{align} $$

Efficiency

• The efficiency, \(\varepsilon\), is the ratio of the work obtained to the heat provided by the fuel, \(q_1\) $$ \begin{align} \varepsilon &= -\frac{w}{q_1} = \frac{nR(T_{hot}-T_{cold})\ln \frac{V_B}{V_A}}{nRT_{hot}\ln \frac{V_B}{V_A}} \\ &= \frac{T_{hot}-T_{cold}}{T_{hot}} \end{align} $$

The Otto Cycle

• Idealized version of the operation of an internal combustion engine
• Instead of two temperature reservoirs, we use two material reservoirs
  • Fuel supply line
  • Exhaust line
• Stesp of the Otto Cycle
  1. At \(P_A\) and \(T_A\), we ignite the compressed fuel at volume \(V_A\)
  2. The pressurized combustion products expand adiabatically to \(P_C\) and \(V_C\)
  3. Piston stop, gas cools isochorically, pressure drops to \(P_D\)
  4. Adiabatic compression back to \(V_A\)

Otto Cycle Efficiency

• After some math we arrive at an equation for the efficiency of the Otto cycle $$ \begin{align} \varepsilon &= \frac{T_{hot}-T_{cold}}{T_{hot}} \\ &= 1-\left(\frac{V_A}{V_B}\right)^{\frac{C_P-C_V}{C_V}} \end{align} $$
• Note the compression ratio, \(\frac{V_A}{V_B}\)

Maximum Work

• The maximum work is accomplished when the process is reversible
• Therefore, uneven or too rapid combustion reduces the engine’s efficiency
• Combustion rate is a function of the fuel and the igniter
  • Adjusting these has long been used to improve engine efficiency