← | ## T-L02 Phase Changes and Ideal Gas Law | → |

- Test Results (if ready)

- volume (V), [m
^{3}] - pressure (p), [Pa]
- mass (M), [kg]
- temperature (T) [K]: related to thermal energy (E
_{th}) - Derived: ρ = M/V (mass density) [kg m-3]

- N: Number of molecules [-]
- N
_{A}: 6.02 x 10^{23}[mol^{-1}] - n = N/N
_{A}[mol]

- Incompatable assumptions:
- molecules don't touch each other;
- gases in thermal equilibrium (requires contact)

- → Valid away from phase transition boundary (see phase diagram)
- Equation:.
- pV = n R T where R= 8.31 [J mol
^{-1}K^{-1}] - pV = N k
_{B}T where k_{B}=1.nt04 x 10^{-23}[J K^{-1}]

- pV = n R T where R= 8.31 [J mol
- p-V diagrams (p,V → T, only concern Δ) for Quasi-Static Processes
isochoric: V

_{f}= V_{i}isobaric: p

_{f}= p_{i}isothermal: T

_{f}= T_{i}

- Ref: Knight Version 3 Chapter 16
- Ref: Knight Version 4 Chapter 18

- Why are the the quantities such are R and N
_{A}not simple numbers? ANS: This is due to history and the way units were first defined. For example, we could define a new unit of temperature the [bai]. If**10 [bai] = 8.31 [K]**then**R=10**. (Water would freeze at 328.7 bai, and boil at 448 bai. I could then define a scale for daily use T [xiaobai] = T[bai] -328.7 so that the freezing point of water would be 0 [xiaobai] and boiling point would be 1nt06.5 [xiaobai] ) Similarily we could define a new unit of mass, the [white] such that so that Avrogadro's number is 10^{24}! - Why can we only consider quasi-static processes in thermodynamics? ANS: P,V and T are only defined for thermodynamic equilibrium. if things move fast we leave this equilibrium and the state variables are defined. i.e. T is not constant in the system, P is not constant. To draw a path on the pV diagram we need to go slow. See Quasistatic and Reversible Processes (Khan Academy)