Thermodynamic Processes and the First Law

Work Done and Thermodynamic Processes

Work Done: Definition and Sign Convention

• Definition: Work (W) is done when a gas expands or contracts against an external pressure. Mathematically, it's defined as W = -[\int P dV]. This definition represents the work done on the gas. The negative sign is crucial: if the gas expands, dV is positive, so the work done on the gas is negative. If the gas is compressed, dV is negative, so the work done on the gas is positive.

• Expansion (V{final} > V{initial}): If the final volume is greater than the initial volume (V{final} > V{initial}), the work done on the gas is negative. This means the gas is doing work on its surroundings (expanding freely, requiring no effort from an external agent to push it). For example, if we are doing the work, it would be negative work by us.

• Contraction/Compression (V{final} < V{initial}): If the final volume is smaller than the initial volume (V{final} < V{initial}), the work done on the gas is positive. This means an external agent is actively compressing the gas, requiring effort (e.g., pushing down on a piston).

• Graphical Interpretation: The integral [\int P dV] represents the area under the curve on a Pressure-Volume (P-V) diagram. Therefore, the work done on the gas (W) is the negative of the area under the P-V curve. The sign of the work depends on whether the process involves expansion or compression.

Isothermal Process: Constant Temperature

• Restriction: The temperature (T) remains constant throughout the process (T = \text{constant}).

• Implications:

  • Since temperature is constant, the change in temperature is zero (\Delta T = 0).

  • For an ideal gas, internal (thermal) energy (E) is solely a function of temperature. Therefore, if \Delta T = 0, then the change in internal energy is also zero (\Delta E = 0).

• First Law of Thermodynamics: Applying the First Law (\Delta E = Q + W), since \Delta E = 0, we get 0 = Q + W. This implies that the heat exchanged (Q) is the negative of the work done (W = -Q).

• Work Calculation: For an isothermal process involving an ideal gas, the work done on the gas is given by:

  W = -nRT \ln\left(\frac{V{final}}{V{initial}}\right)

• Heat Calculation: From $Q = -W$, the heat exchanged is:

  Q = nRT \ln\left(\frac{V{final}}{V{initial}}\right)

Isometric (Isovolumetric) Process: Constant Volume

• Restriction: The volume (V) remains constant throughout the process (V = \text{constant}).

• Implications:

  • Since volume is constant, the change in volume is zero (dV = 0).

  • Work Calculation: As W = -\int P dV, if dV = 0, then the work done on the gas is zero (W = 0).

• First Law of Thermodynamics: With W = 0, the First Law (\Delta E = Q + W) simplifies to \Delta E = Q .

• Heat/Internal Energy Calculation: The heat added or removed from the gas (which equals the change in internal energy) is given by:

  Q = \Delta E = nCV \Delta T where CV is the molar specific heat at constant volume.

• P-V Diagram: On a P-V diagram, an isometric process is represented by a vertical line, as volume does not change while pressure can.

Free Expansion

• Definition: A process where a gas expands into a vacuum without any external forces acting on it. Example: Opening a door to a vacuum-filled hallway and letting air expand freely.

• Restrictions/Implications:

  • Work: No external force means no work is done on or by the gas (W = 0).

  • Heat: Typically, free expansion is considered to occur instantaneously within a thermally insulated system, meaning no heat is exchanged (Q = 0).

  • First Law of Thermodynamics: With W = 0 and Q = 0, the First Law (\Delta E = Q + W) yields \Delta E = 0 .

  • Temperature Change: Since \Delta E = 0 and thermal energy for an ideal gas depends only on temperature, it follows that \Delta T = 0 . Although the gas occupies a larger volume and its pressure changes, the overall thermal energy and temperature remain constant.

Cyclic Process

• Definition: A process in which a system starts from an initial state, undergoes a series of changes, and returns to its exact initial state.

• Implications for State Functions:

  • Temperature: Temperature is a state function, meaning its value depends only on the current state of the system, not on the path taken. Since the system returns to its initial state, the total change in temperature for a complete cycle is zero (\Delta T_{total} = 0).

  • Internal Energy: Internal energy is also a state function. Therefore, the total change in internal energy for a complete cycle is zero (\Delta E_{total} = 0).

• First Law of Thermodynamics for a Cycle: From \Delta E{total} = Q{total} + W{total} and \Delta E{total} = 0, we get Q{total} = -W{total} . This means the net heat absorbed equals the net work done by the system (or net heat added equals negative net work done on the system).

• Total Work Calculation (P-V Diagram): The total work done on the gas in a cyclic process is equal to the negative of the area enclosed by the cycle on the P-V diagram.

  • If the cycle proceeds in a clockwise direction, the system does positive work, meaning W_{total} (work on the gas) is negative. The expansion phase (larger volume) typically has a larger area under its curve than the compression phase (smaller volume) when moving clockwise.

  • If the cycle proceeds in a counter-clockwise direction, the system does negative work, meaning W_{total} (work on the gas) is positive. The compression phase (smaller volume) typically has a larger area Magnitude under its curve than the expansion phase (larger volume) when moving counter-clockwise.

• Calculating Work for Individual Legs: For portions of the cycle with constant volume, work is zero. For portions with constant pressure, work is W = -P \Delta V. For other curves (e.g., isothermal), specific formulas apply.

Adiabatic Process: No Heat Transfer

• Restriction: No heat is allowed to enter or leave the system (Q = 0). This implies the process occurs rapidly or the system is perfectly thermally insulated (e.g., a thermos).

• First Law of Thermodynamics: With Q = 0, the First Law (\Delta E = Q + W) simplifies to \Delta E = W . This means any work done on the gas directly changes its internal energy, and vice-versa.

• Changing Variables: Unlike isothermal or isobaric processes, in an adiabatic process, all three variables (pressure P, volume V, and temperature T) typically change.

• Adiabatic Relation between P and V: A key relationship for an adiabatic process involving an ideal gas is: PV^\gamma = \text{constant} where \gamma (gamma) is the adiabatic constant or ratio of specific heats, defined as the ratio of the molar specific heat at constant pressure (CP) to the molar specific heat at constant volume (CV): \gamma = \frac{CP}{CV}

  • For a monatomic ideal gas (e.g., He, Ne): CV = \frac{3}{2}R, CP = \frac{5}{2}R, so \gamma = \frac{5/2 R}{3/2 R} = \frac{5}{3} \approx 1.67 .

  • For a diatomic ideal gas (e.g., N$2$, O$2$): CV = \frac{5}{2}R, CP = \frac{7}{2}R, so \gamma = \frac{7/2 R}{5/2 R} = \frac{7}{5} = 1.40 .

  • Since \gamma > 1, the adiabatic curve (called an adiabat) on a P-V diagram is always steeper than an isothermal curve. This means pressure falls off faster with volume in an adiabatic expansion compared to an isothermal expansion.

• Alternative Adiabatic Relations: Using the ideal gas law (PV = nRT), other relations can be derived:

  • TV^{\gamma-1} = \text{constant}

  • P^{1-\gamma}T^\gamma = \text{constant}

• Work Calculation: The work done on the gas during an adiabatic process is given by:

  W = \frac{P{final}V{final} - P{initial}V{initial}}{1 - \gamma}

  This formula can also be expressed in terms of temperature:

  W = nCV(T{final} - T{initial}) (since \Delta E = nCV \Delta T and W = \Delta E)

• Real-world Examples: Sound waves propagate adiabatically through air. The early expansion of the universe is also well-approximated as an adiabatic process.

Summary Table of Thermodynamic Processes

Example: Cyclic Process Calculation

Consider a cyclic process A \to B \to C \to A on a P-V diagram, with $n = 0.75$ moles of a diatomic gas. Given data:

• Point A: PA = 3.2 \times 10^3 \text{ Pa}, VA = 0.21 \text{ m}^3

• Point B: PB = 1.2 \times 10^3 \text{ Pa}, VB = V_A = 0.21 \text{ m}^3

• Process C \to A is isothermal.

Goal: Find Q, W, and \Delta E for each process (A \to B, B \to C, C \to A) and for the total cycle.

Step 1: Determine all P, V, T values at each point.

• At Point A:

  Using ideal gas law (PA VA = n R TA): TA = \frac{PA VA}{nR} = \frac{(3.2 \times 10^3 \text{ Pa})(0.21 \text{ m}^3)}{(0.75 \text{ mol})(8.31 \text{ J/(mol K)})} \approx 108 \text{ K}

• At Point B:

  We know PB = 1.2 \times 10^3 \text{ Pa} and VB = VA = 0.21 \text{ m}^3. TB = \frac{PB VB}{nR} = \frac{(1.2 \times 10^3 \text{ Pa})(0.21 \text{ m}^3)}{(0.75 \text{ mol})(8.31 \text{ J/(mol K)})} \approx 40 \text{ K}

• At Point C:

  From the diagram, PC = PB = 1.2 \times 10^3 \text{ Pa}.

  Since C \to A is an isothermal process, TC = TA = 108 \text{ K}.

  VC = \frac{n R TC}{P_C} = \frac{(0.75 \text{ mol})(8.31 \text{ J/(mol K)})(108 \text{ K})}{(1.2 \times 10^3 \text{ Pa})} \approx 0.56 \text{ m}^3

Summary of States:

• A: PA = 3.2 \times 10^3 \text{ Pa}, VA = 0.21 \text{ m}^3, T_A = 108 \text{ K}

• B: PB = 1.2 \times 10^3 \text{ Pa}, VB = 0.21 \text{ m}^3, T_B = 40 \text{ K}

• C: PC = 1.2 \times 10^3 \text{ Pa}, VC = 0.56 \text{ m}^3, T_C = 108 \text{ K}

Step 2: Calculate Q, W, \Delta E for each process.

• Process A \to B (Isochoric - Constant Volume):

  • V is constant, so dV = 0. Thus, W_{AB} = 0 .

  • From the First Law, \Delta E{AB} = Q{AB} + W{AB} = Q{AB} .

  • For a diatomic gas, C_V = \frac{5}{2}R .

  • Q{AB} = nCV \Delta T{AB} = (0.75 \text{ mol})(\frac{5}{2})(8.31 \text{ J/(mol K)})(TB - T_A)

    Q_{AB} = (0.75)(\frac{5}{2})(8.31)(40 - 108) \approx -1060 \text{ J}

  • Therefore, \Delta E_{AB} = -1060 \text{ J} .

  • Interpretation: Temperature falls, so gas loses heat to surroundings. Negative $Q$ means heat leaves the system.

• Process B \to C (Isobaric - Constant Pressure):

  • P is constant (PB = PC = 1.2 \times 10^3 \text{ Pa}).

  • Work done on the gas: W{BC} = -PB (VC - VB)

    W_{BC} = -(1.2 \times 10^3 \text{ Pa})(0.56 \text{ m}^3 - 0.21 \text{ m}^3) \approx -420 \text{ J}

  • For a diatomic gas, CP = CV + R = \frac{5}{2}R + R = \frac{7}{2}R .

  • Heat exchanged: Q{BC} = nCP \Delta T{BC} = (0.75 \text{ mol})(\frac{7}{2})(8.31 \text{ J/(mol K)})(TC - T_B)

    Q_{BC} = (0.75)(\frac{7}{2})(8.31)(108 - 40) \approx 1480 \text{ J}

  • From the First Law: \Delta E{BC} = Q{BC} + W_{BC} = 1480 \text{ J} + (-420 \text{ J}) = 1060 \text{ J} .

  • Interpretation: Temperature rises, so heat is added (positive $Q$). Gas expands, doing work on surroundings (negative $W$).

• Process C \to A (Isothermal - Constant Temperature):

  • T{CA} = \text{constant} (TC = T_A = 108 \text{ K}).

  • Since temperature is constant, \Delta E_{CA} = 0 .

  • From the First Law, Q{CA} = -W{CA} .

  • Work done on the gas: W{CA} = -nRTA \ln\left(\frac{VA}{VC}\right)

    W_{CA} = -(0.75 \text{ mol})(8.31 \text{ J/(mol K)})(108 \text{ K}) \ln\left(\frac{0.21 \text{ m}^3}{0.56 \text{ m}^3}\right) \approx 660 \text{ J}

  • Therefore, Q_{CA} = -660 \text{ J} .

  • Interpretation: Gas is compressed ($VA < VC$), so positive work is done on the gas. Since $Q = -W$, heat leaves the system ($Q$ is negative) to maintain constant temperature during compression.

Step 3: Calculate Total Q, W, \Delta E for the entire cycle.

• Total Heat (Q_{total}):

  Q{total} = Q{AB} + Q{BC} + Q{CA} = -1060 \text{ J} + 1480 \text{ J} + (-660 \text{ J}) = -240 \text{ J}

• Total Work (W_{total}):

  W{total} = W{AB} + W{BC} + W{CA} = 0 \text{ J} + (-420 \text{ J}) + 660 \text{ J} = 240 \text{ J}

• Total Internal Energy Change (\Delta E{total}): One way: \Delta E{total} = Q{total} + W{total} = -240 \text{ J} + 240 \text{ J} = 0 \text{ J} (This confirms Q{total} = -W{total}).

  Another way (sum of individual changes): \Delta E{total} = \Delta E{AB} + \Delta E{BC} + \Delta E{CA} = -1060 \text{ J} + 1060 \text{ J} + 0 \text{ J} = 0 \text{ J}.

  • Interpretation: For any cyclic process, the total change in internal energy must be zero, as the system returns to its initial state. This result serves as a good check for the calculations.

Summary: This example demonstrates how to methodically apply the definitions and formulas for different thermodynamic processes, using the First Law of Thermodynamics and the ideal gas law to track the changes in heat, work, and internal energy.