Lecture 5 First Law of Thermodynamics

You can’t get something for nothing. Nothing is for free. We will discuss these statements later…

Thermodynamic System Consider a closed system (e.g., a parcel of air). It has internal energy (“u”) = energy due to molecular kinetic and potential energies. Suppose some energy (dq) is added to the system. Example: via radiation from the sun What happens?

Thermodynamic System Some of the energy goes into work done (dw) by the system against its surroundings. Example: expansion What’s left is a change in internal energy. du = dq – dw Conservation of Energy principle “nothing is for free”

Thermodynamic System System Environment Heat System can exchange energy with environment via heat flow.

Thermodynamic System In addition, system can do work on environment or vice versa. Example: expansion

First Law, General Form dU = dQ – dW dU = change in internal energy of system dQ = heat exchanged with environment dQ > 0  heat flowing into system dQ > 0  heat flowing into system dW = work done by or on system dW > 0  system is doing work dW > 0  system is doing work

Ideal Gas Consider a system consisting of an ideal gas in a cylinder. Cylinder has a piston, which allows volume to be changed.

Cylinder weights more weight, more pressure Gas Walls of cylinder: 1) perfect heat conductors 2) perfect insulators

Cases Case 1: Walls of cylinder are perfect conductors  heat can freely flow between system and environment  heat can freely flow between system and environment  in equilibrium, temperature of system must equal temperature of environment  in equilibrium, temperature of system must equal temperature of environment Case 2: Walls of cylinder are perfect insulators  no heat flow between system and environment  no heat flow between system and environment  temperatures of system and environment need not be equal  temperatures of system and environment need not be equal

Work (Qualitative) System does work on environment (dW > 0) if gas expands (Piston is pushed upward.) (Piston is pushed upward.) Work is done on system (dW < 0) if gas contracts (Piston is pushed downward.) (Piston is pushed downward.)

Work (Quantitative) Suppose piston is pushed upward a distance dx.

Work (Quantitative) Suppose piston is pushed upward a distance dx. dx Area of piston = A Pressure force = p  A dW = p  A  dx = pdV

First Law, Ideal Gas Usually, we are interested in energy per unit mass Divide both sides by m Define u = U/m; q = Q/m;  = V/m

First Law, Ideal Gas (per unit mass)

Internal Energy For an ideal gas, u is a function of T only u is an increasing function of T u is an increasing function of T Change in internal energy depends only on change in temperature. Doesn’t depend on the way in which that temperature change is accomplished. Doesn’t depend on the way in which that temperature change is accomplished. du = k  dT, where k is a constant du = k  dT, where k is a constant (Value of k will be determined shortly.) (Value of k will be determined shortly.)

Heat Capacity, C For a gas, C depends on the particular process. C v = heat capacity at constant volume (dV = 0) C p = heat capacity at constant pressure (dp = 0) SI units: J  K -1 Amount of heat required to change a “body’s” temperature by a given amount

Specific Heat Specific heat is heat capacity per unit mass c v (lower case) = specific heat at constant volume c p = specific heat at constant pressure Specific heat is the heat energy needed to raise the temperature of a unit mass of a substance by one degree.

Specific Heat Suppose we add some thermal energy (dq) to a unit mass of a substance like air, water, soil. We expect T(substance) to increase How much? We can define Specific Heat as Heat added Temp change Constant volumeConstant pressure

Specific Heat of Dry Air c v = 717 J  kg -1  K -1 c p = 1004 J  kg -1  K -1 Note: c p – c v = 287 J  kg -1  K -1 Look familiar? Look familiar? c p – c v = R d c p – c v = R d (Not a coincidence!) (Not a coincidence!)

Constant Volume Processes 0

Internal Energy Change The following is always true: (3.40) W&H

First Law for Ideal Gas, Re-written (1)

Constant-Pressure Processes Go back to ideal-gas law: Take differential of both sides:

But,…  (2)

Substitute (2) into (1)

Simplify Constant-pressure process  (3)

But, … For a constant-pressure process,

Compare and Result:

Rewrite (3) We will use this in the next lecture. (4)