Thermodynamics Notes

System and Surroundings

• System: the object or quantity being observed or its properties that are observed/calculated.

• Surroundings: everything else outside the system.

• Universe: system + surroundings.

• Practical implication: thermodynamic analysis focuses on transfer of energy (and/or matter) between the system and its surroundings.

Types of Systems

• Open system: exchange of matter and energy between the system and surroundings.

• Closed system: no exchange of matter with surroundings; energy exchange is possible.

• Isolated system: no exchange of matter or energy with surroundings.

Thermodynamic Process and Entropy

• Thermodynamic process: method by which a system is changed from one state to another.

• Entropy: a measure of the degree of disorder in a system.

First Law of Thermodynamics (in words and equations)

• Change in internal energy of a system

  • \Delta U = Q - W

  • Sign convention:

  • Q: Heat flow into the system; +Q means heat added to the system, −Q means heat leaving the system.

  • W: Work done by the system; +W means work done by the system on the surroundings, −W means work done on the system.

  • If the signs are interpreted as in the table below:

    • + Sign for Q corresponds to heat flow into the system

    • + Sign for W corresponds to work done by the system

    • + Sign for ΔU corresponds to internal energy increase

• Sign convention table (summary)

  • Q\quad\text{(heat flow into the system)}

  • +Q: heat flows into the system

  • −Q: heat flows out of the system

  • W\quad\text{(work done by the system)}

  • +W: work done by the system

  • −W: work done on the system

  • \Delta U\quad\text{(internal energy)}

  • +\Delta U: internal energy increase

  • −\Delta U: internal energy decrease

Special Cases of the 1st Law

• Adiabatic process: Q = 0\quad\Rightarrow\quad \Delta U = -W

• Constant Volume: W = 0\quad\Rightarrow\quad \Delta U = Q

• Closed Cycle: \Delta U = 0\quad\Rightarrow\quad Q = W

• Free Expansion: Q = 0,\; W = 0\quad\Rightarrow\quad \Delta U = 0

Thermodynamic Process: Mechanical Work

• State variables describe the state of a system: pressure (P), temperature (T), volume (V), number of moles (n), internal energy (U), entropy (S).

• These state variables describe the state at an instant, but not how the system arrived there; heat and work are not state functions.

Mechanical Work with a Piston (P-V system)

• Consider a gas in a cylinder with a movable piston of cross-sectional area A.

• Gas exerts a force F = P A on the piston.

• If the piston moves a small distance \Delta x, the work done by the gas is:

  • \Delta W = F \Delta x = P A \Delta x = P (A \Delta x) = P \Delta V

• For a large displacement broken into small steps \Delta xj with volume changes \Delta Vj = A \Delta x_j, the total work is:

  • WT = \sumj \Delta Wj = \sumj P \Delta V_j

• If P varies during the process, the total work is the area under the P-V path, i.e.:

  • W_T = \int P \mathrm{d}V

Work Done by an Ideal Gas at Constant Pressure (Isobaric)

• Isobaric process: pressure remains constant; the PV-diagram line is horizontal.

• Work done equals the area under the line:

  • WT = \int{Vi}^{Vf} P \, dV = P (Vf - Vi)

  • Since P V = n R T for an ideal gas, can also be written as WT = P Vf - P Vi = P (Vf - V_i)

Work Done by an Ideal Gas at Constant Temperature (Isothermal)

• Isothermal process: temperature remains constant.

• General expression for work during a volume change:

  • WT = \int{Vi}^{Vf} P \, dV

• For an ideal gas, P V = n R T, so with constant T, pressure is P = \dfrac{n R T}{V}.

• Substitution gives:

  • WT = \int{Vi}^{Vf} \dfrac{n R T}{V} \, dV = n R T \ln\left(\dfrac{Vf}{Vi}\right)

• Using the relation Pi Vi = Pf Vf = n R T, one may also write:

  • WT = n R T \ln\left(\dfrac{Pi}{P_f}\right)

Change in Entropy

• Reversible process (change in entropy due only to internal energy change):

  • \Delta S = \frac{\Delta Q_{rev}}{T}

  • If temperature remains approximately constant, this holds as an approximation.

  • Conditions allowing this approximation include:

  • heat capacity is large (reservoir-like)

  • the amount of heat flow Q is small

  • the system does work W = Q so that \Delta U = 0

• Heat transfer between two systems at different temperatures (finite heat transfer):

  • If hot reservoir at Th and cold reservoir at Tc exchange heat amount Q, the total entropy change is:

  • \Delta S = - \frac{Q}{Th} + \frac{Q}{Tc}

  • Since Th > Tc, \dfrac{Q}{Th} < \dfrac{Q}{Tc}, so the total entropy change can be positive or negative depending on the net heat exchange.

• When the temperature changes during the process (\Delta T \neq 0):

  • The temperature interval is divided into small intervals; entropy change is the sum over intervals, taking the limit as the interval goes to zero:

  • \Delta S = \Delta Sf = Sf - Si = \int{Ti}^{Tf} \dfrac{\mathrm{d}Q}{T}

  • In this framework, Q is the energy transferred as heat and T is the temperature in Kelvin.

• SI unit of entropy: J K⁻¹ (or cal K⁻¹ in other unit systems).

The Second Law of Thermodynamics

• The total entropy of a system plus its surroundings never decreases:

  • \Delta S_{total} \ge 0

• If \Delta S_{total} > 0, the process is naturally occurring (irreversible).

• If \Delta S_{total} = 0, the process is reversible (idealized, equilibrium, infinitely slow).

• If \Delta S_{total} < 0, the process cannot occur.

Example: Coffee Cooling Problem (Entropy of Coffee and Surroundings)

• Given: coffee mass m = 0.25\text{ kg}, specific heat capacity c = 4186\;\text{J kg}^{-1}\text{K}^{-1} (water-like), initial temperature Ti = 373\text{ K}, final temperature Tf = 293\text{ K}.

• Change in entropy of the coffee:

  • \Delta SC = m c \ln\left(\frac{Tf}{T_i}\right)

  • \Delta S_C = 0.25 \times 4186 \times \ln\left(\frac{293}{373}\right) \approx -2.53\times 10^{2}\ \text{J K}^{-1}

• Entropy change in the surroundings (constant surroundings temperature, assumed):

  • The coffee loses heat: Q = m c (Tf - Ti) = m c \Delta T with \Delta T = -80\text{ K} (cooling). The surroundings gain heat: \Delta S{sur} = \frac{Q}{T{sur}} = \frac{m c (Ti - Tf)}{T_{sur}} = \frac{0.25 \times 4186 \times 80}{293} \approx 2.86\times 10^{2} \ \text{J K}^{-1}

• Entropy change for coffee + surroundings (universe):

  • \Delta S{total} = \Delta SC + \Delta S_{sur} \approx -2.53\times 10^{2} + 2.86\times 10^{2} \approx +3.31\times 10^{1} \ \text{J K}^{-1}

Heat Engines: Efficiency

• Efficiency of a heat engine:

  • \eta = \frac{\text{Net work output}}{\text{Heat input from hot reservoir}} = \frac{W{net}}{QH}

• From the First Law, for a heat engine: W{net} = QH - QC\;\Rightarrow\; \eta = 1 - \frac{QC}{Q_H}

Entropy in a Heat Engine

• Entropy changes:

  • \Delta SH = -\frac{QH}{T_H}

  • \Delta SC = +\frac{QC}{T_C}

  • Engine and surroundings: \Delta SE = 0,\quad \Delta SR = 0

  • Total: \Delta S{total} = -\frac{QH}{TH} + \frac{QC}{TC} = \frac{QC}{TC} - \frac{QH}{T_H}

Carnot’s Principle and Carnot Engine

• No irreversible engine between two reservoirs can have greater efficiency than a reversible engine operating between the same two reservoirs.

• All reversible engines between the same two reservoirs have the same efficiency.

• For a reversible engine, the ratio of the heats equals the ratio of the temperatures:

  • \frac{QH}{QC} = \frac{TH}{TC}\,.

• Carnot cycle implies total entropy change is zero:

  • \Delta S_{total} = 0

  • Therefore: \frac{QH}{QC} = \frac{TH}{TC}

• Efficiency of a Carnot engine:

  • \eta = \frac{W{net}}{QH} = 1 - \frac{QC}{QH} = 1 - \frac{TC}{TH}

Refrigerators and Heat Pumps

• Refrigerator Coefficient of Performance (COP_R):

  • \text{COP}R = \frac{\text{Desired effect}}{\text{Work input}} = \frac{QC}{W_{net}}

  • Using First Law: W{net} = QH - QC\Rightarrow \text{COP}R = \frac{QC}{QH - QC} = \frac{1}{\dfrac{QH}{Q_C} - 1}

• Heat Pump Coefficient of Performance (COP_HP):

  • \text{COP}{HP} = \frac{\text{Desired effect}}{\text{Work input}} = \frac{QH}{W_{net}}

  • Using First Law: W{net} = QH - QC\Rightarrow \text{COP}{HP} = \frac{QH}{QH - QC} = \frac{1}{1 - \dfrac{QC}{Q_H}}

• Relationship between COPHP and COPR:

  • \text{COP}{HP} = \text{COP}R + 1

• Conceptual layout: warm environment is at temperature TH> TC; cold space is at T_C; heat transfer involves heat flows between reservoirs and the device performing work.

Summary of Key Equations (LaTeX)

• First Law: \Delta U = Q - W

• Isobaric work: W = P (Vf - Vi)

• Isothermal work for ideal gas: W = n R T \ln\left(\dfrac{Vf}{Vi}\right) = n R T \ln\left(\dfrac{Pi}{Pf}\right)

• Entropy change (reversible): \Delta S = \frac{\Delta Q{rev}}{T} = \int \frac{\mathrm{d}Q{rev}}{T}

• Entropy change for heat exchange between reservoirs: \Delta S = -\frac{Q}{Th} + \frac{Q}{Tc}

• Second Law (total): \Delta S_{total} \ge 0

• Carnot efficiency: \eta{Carnot} = 1 - \frac{TC}{T_H}

• Entropy changes in a heat engine: \Delta SH = -\frac{QH}{TH},\quad \Delta SC = \frac{QC}{TC},\quad \Delta SE = 0,\quad \Delta SR = 0

• Refrigerator COP: \text{COP}R = \frac{QC}{QH - QC}

• Heat pump COP: \text{COP}{HP} = \frac{QH}{QH - QC}

• Relationship: \text{COP}{HP} = \text{COP}R + 1

Real-World and Conceptual Takeaways

• Distinguish system vs surroundings in any problem; keep track of what crosses system boundaries.

• Not all energy transfers are state functions; only state variables describe a state at an instant.

• The Second Law introduces irreversibility and the concept of entropy as a measure of disorder and energy dispersal.

• Carnot's principle sets the upper limit on efficiency for engines operating between two reservoirs and is realized only by ideal reversible cycles.

• COPs provide practical measures of performance for refrigerators and heat pumps, showing trade-offs between energy flows and useful effects.