First Law Flashcards
First Law of Thermodynamics
dU = dQ + dW
U = internal energy Q = heat W = work
Helmholtz Energy
ΔF = ΔU + TΔS
How does thermodynamics describe the world?
separates the world into the system and the surroundings
Isolated System
-doesn’t exchange energy or matter with the surroundings
Closed System
-exchanges energy but not matter with the surroundings
Open System
-exchanges energy and matter with the surroundings
State Variables
T, V, P, U
State Functions
-functions of state variables
Extensive Variables
-depend on the size of the system, e.g. U, Nk
Intensive Variables
-specify a local property independent of the size of a system e.g. T, P
Zeroth Law of Thermodynamics
- if 2 systems are in thermal equilibrium with a third then they are also in equilibrium with each other
- there is a single property, temperature, that is common to all three systems and serves to indicate that they are in thermal equilibrium
First Law
-when a system undergoes a change of state, the sum of the different energy changes (heat exchanges, work, etc.) is independent of the manner of the transformation and depends only on the initial and final states of the system:
B,A ∫ dU = Ub - Ua
OR for a cyclic process:
∮ dU = 0
Perpetual Motion of the First Kind
- it is not possible either by mechanical, thermal, chemical or other devices to obtain perpetual motion
- i.e. it is impossible to construct an engine that will work in a cycle and produce continuous work or kinetic energy from nothing
First Law
Closed System
dU = dQ + dW
dU = energy exchanged with thermal reservoir
First Law
Open System
dU = dQ + dW + dUmatter
Types of Work
Extension, Surface Expansion, Externsion
Extension: dW = Fdl F = tension, l = length Surface Expansion: dW = γdA γ = surface tension, A = area Expansion: dW = PdV P = external pressure, V = volume
-all are a product of an intensive factor and an extensive factor
Types of Work
Electrical, Magnetic
Electrical: dW = ϕdQ
ϕ = potential difference, Q = charge
Magnetic: dW = -BdM
B = magnetic field, M = change in magentic dipole moment
Work Done During any Infinitesimal Change in Volume
dW = -Pex dV
Pex = external pressure
Reversible Change
Pex is always matched to Pint
Irreversible Change
Pex is less than the internal pressure
Pex
Isothermal Expansion of an Ideal Gas
T = constant dU = dQ + dW = 0 dQ = -dW
-energy that enters the system as heat, leaves as work
Adiabatic Expansion of an Ideal Gas
dQ = 0 dU = dQ + dW = dW
-change in internal energy is equal to the work done
Ideal Gas Assumptions
1) particles do not interact (except through collisions), it doesn’t matter how close two particles get their potential energy is the same, this breaks down at high pressure as in reality all substance experience van der Waal’s interaction
2) particles do not take up any volume
Calculating Heat Capacity
Finding an Expression for Heat
dU = dQ + dW dQ = dU - dW =dU + PdV = δU/δT|v dT + δU/dV|t dV + PdV =δU/δT|v dT + (δU/δV|t + P)dV
Calculating Heat Capacity
Constant Volume
-start with the expression for heat: dQ = δU/δT|v dT + (δU/δV|t + P)dV -at constant volume, dV=0 so: dQ = δU/δT|v dT Cv = dQ/dT Cv = δU/δT|v
Calculating Heat Capacity
Constant Pressure
start with the expression for heat:
dQ = δU/δT|v dT + (δU/δV|t + P)dV
Cp = dQ/dT |p
Cp = δU/δT|v + (δU/δV|t + P) dV/dT
Calculating Heat Capacity
Difference in Cp and Cv for all substances
Cp - Cv = δU/δT|v + (δU/δV|t + P) dV/dT - δU/δT|v
= (δU/δV|t + P) dV/dT
Classical Equipartition Theorem
-every degree of freedom contributes 1/2kT to the internal energy of the system
Gamma
γ = Cp/Cv
for a monoatomic gas γ=5/3
for a diatomc gas γ=7/5
Application of the First Law - Speed of Sound
-due to the rapid nature of pressure changes that propagate sound, hardly any heat is exchanged - appro. adiabatic
-speed of sound in medium depends on its bulk modulus and density Vs = √(B/ρ)
-bulk modulus B = γP
-for an ideal gas:
P = nRT/V = nM/V * RT/M = ρ RT/M
B = γρRT/M
Vs = √(B/ρ) = √(γRT/M)
Calculating Heat Capacity
Difference in Cp and Cv for an Ideal Gas
starting with the general expression:
Cp - Cv = (δU/δV|t + P) dV/dT
-for an ideal gas:
PV=nRT so V = nRT/P and dV/dT = nR/P
Cv - Cp = (δU/δV|t + P) * nR/P
-for an ideal gas, internal energy doesn’t depend on volume so, δU/δV|t=0
Cv - Cp = P * nR/P = nR
Reversible or Quasistatic Processes
- idealised processes that take place infinitely slowly in such a way that the system is always in equilibrium at every stage
- no real thermodynamic changes are quasistatic due to friction etc.
- BUT properties of state functions (that they only depend on the initial and final state not the path taken) allow us to define a quasistatic path between to states in order to calculate thermodynamic potentials of the initial and final states that will be the same for the actual path taken
