Transcript Lecture 5
Lecture 5 First Law of Thermodynamics 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 Heat Environment System 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 dW = work done by or on system 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 Walls of cylinder: Gas 1) perfect heat conductors 2) perfect insulators Cases Case 1: Walls of cylinder are perfect conductors heat can freely flow between system and 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 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.) Work is done on system (dW < 0) if gas contracts (Piston is pushed downward.) Work (Quantitative) Suppose piston is pushed upward a distance dx. Work (Quantitative) Suppose piston is pushed upward a distance dx. Area of piston = A Pressure force = pA dx dW = pAdx = pdV First Law, Ideal Gas dU dQ pdV 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) du dq pd Internal Energy For an ideal gas, u is a function of T only 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. du = kdT, where k is a constant (Value of k will be determined shortly.) Heat Capacity, C dQ C dT SI units: JK-1 Amount of heat required to change a “body’s” temperature by a given amount For a gas, C depends on the particular process. Cv = heat capacity at constant volume (dV = 0) Cp = heat capacity at constant pressure (dp = 0) Specific Heat Specific heat is heat capacity per unit mass cv (lower case) = specific heat at constant volume cp = 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? Heat added dq We can define Specific Heat as dT Temp change dq cv dT v dq cp dT p Constant volume Constant pressure Specific Heat of Dry Air cv = 717 Jkg-1K-1 cp = 1004 Jkg-1K-1 Note: cp – cv = 287 Jkg-1K-1 Look familiar? cp – cv = Rd (Not a coincidence!) Constant Volume Processes du dq pd du dq cv d T 0 Internal Energy Change The following is always true: du cv d T (3.40) W&H First Law for Ideal Gas, Re-written cv dT dq pd (1) Constant-Pressure Processes Go back to ideal-gas law: p RT Take differential of both sides: d ( p ) RdT But,… d ( p ) pd dp pd RdT dp (2) Substitute (2) into (1) pd RdT dp cv dT dq pd cv dT dq RdT dp Simplify cv RdT dq dp (3) Constant-pressure process dp 0 dq cv RdT But, … For a constant-pressure process, dq c pdT Compare dq cv RdT and dq c pdT Result: c p cv R Rewrite (3) c pdT dq dp We will use this in the next lecture. (4)