Internal Energy (example)

Transcription

1 Internal Energy (example) A bucket of water KEs: translational: rotational: vibrational: PEs: within molecules: between rest on the table molecular bonds dipole-dipole interactions

2 Internal Energy for an Ideal Gas (monoatomic) Recall in an Ideal Gas No intermolecular forces We can explicitly calculated its Internal Energy U: 3 translational degrees of freedom only! N molecules each contributes ½ kt 1 3 U 3N kt NkT 2 2 (Ideal Gas monatomic) The Internal Energy for an Ideal Gas U is a function of T ONLY! (Experimentally, this is also shown to be the case for most diluted real gases!)

3 The 1 st Law of Thermodynamics We have seen that both heat and work can change the internal energy of the system. In particular, Heat Work +W +Q Piston pushing up on gas Heat enters the system, U increases ( U > 0) Work done by the system, U decreases ( U < 0)

4 The 1 st Law of Thermodynamics Putting these two mechanisms together, we can combine them into one mathematical statement for the change in the internal energy U (total energy) in any processes, U Q W (1 st Law of Therm.) This is a generalization of the principle of conservation of energy to include energy transfer through heat as well as through mechanical means (work).

5 The 1 st Law of Thermodynamics Important Property of U Although both Q and W are path/process dependent U is a state variable and it depends on the initial and final states only. Specially, for an Ideal Gas, U is a function of T only. U is independent of path since U represents the change in the total internal energy of the system

6 1 st Law of Thermodynamics U can be positive, negative,and zero.

7 Special Processes 1. Isolated Systems (no interactions with surrounding) Since Q = W = 0, U 0 U remains a constant in an isolated system. 2. Cyclic Processes (starting state = end state) Since U is a state variable, U = 0. 1 st Law gives, U Q W 0 P Q W (= area encircled by curve) V

8 Special Processes 3. Isochoric (constant V): NO volume change NO work (W=0) W 0 U Q W U Q 4. Isobaric (constant P) Since P is a constant, dw = PdV can be integrated easily and gives, W P( V V ) 2 1 U Q P V 2 V1

9 Example: A Thermodynamical Cycle Example 19.4(2) Given: Q ab = 150 J Q bd = 600 J Find: U ab =? U abd =? Q acd =? note

10 Special Processes 5. Isothermal (constant T) For an Ideal Gas, U(T) depends only on T U = 0! U 0 U Q W Q W Recall, work done by an isothermal process in an Ideal Gas: V2 V2 nrt V 2 ln V V V V 1 W PdV dv nrt 1 1

11 Special Processes 6. Adiabatic (no heat exchange, Q = 0): Q 0 U Q W U W Note: The compression stroke in an internal combustion engine is quick and it can be well approximated by an adiabatic process. (will come back to this example later) D fire piston

12 Summary Four Reversible (Quasi-static) processes: 1. Adiabatic (Q=0) 2. Isochroic ( V=0) 3. Isobaric ( P=0) 4. Isothermal ( T=0) 1 st Law for infinitesimal changes: du dq dw dq PdV

13 C p and C v for an Ideal Gas Two different ways to change dt=t 2 -T 1 : Process a: Constant V a a b b a dq a nc dt v

14 C p and C v for an Ideal Gas For the same dt : Process b: Constant P a a b b dq b nc dt Which is bigger: Q or Q, C or C? a b v P P b

15 C p and C v for an Ideal Gas First let consider the constant volume process (a): No work done by/on gas W = 0 Then, 1 st law gives, du dq nc dt a a v Now, for the constant pressure process (b) with the same dt: We have dw = PdV and dq b = nc p dt (by definition) Substitute them into the 1 st law again gives, dub dqb dw nc pdt PdV