• Shuffle
    Toggle On
    Toggle Off
  • Alphabetize
    Toggle On
    Toggle Off
  • Front First
    Toggle On
    Toggle Off
  • Both Sides
    Toggle On
    Toggle Off
  • Read
    Toggle On
    Toggle Off
Reading...
Front

Card Range To Study

through

image

Play button

image

Play button

image

Progress

1/35

Click to flip

Use LEFT and RIGHT arrow keys to navigate between flashcards;

Use UP and DOWN arrow keys to flip the card;

H to show hint;

A reads text to speech;

35 Cards in this Set

  • Front
  • Back

How to convert from Celsius to Fahrenheit?

Tf = 9/5Tc +32

Internal energy (U)

Thermal Energy due to the motion of all the particles (KE)

Average Kinetic Energy

3/2 kB T

kB

Boltzmann's constant 1.38e-23 J/K

Internal energy of a gas equation

3/2 x molecules x kB x T


3/2 x n x Avogadro x kB x T


3/2 x n x R x T

Root mean squared speed equation

Vrms = sqrtkB*T/m or sqrtRT/M

Linear expansion

Δl = αlΔT

Volumetric expansion

ΔV = βVΔT

Heat (Q)

can be transferred between objects always from hot to cold

+Q

heat flowing into an object

-Q

heat flowing out of an object

Specific heat capacity

amount of heat required to change 1kg of material by 1 degree C or K

Rate of heat transfer (H)

Q/t = k(material)A(area)ΔT/l(length)

Latent heat of transformation (L)

energy required to change 1kg of a material

Lf

Latent heat of fusion (solid to liquid)

Lv

Latent heat of vaporization (liquid to gas)

Thermodynamics

the study of heat transfer or work done

Zeroth Law

Ta= Tb and Tb=Tc then Ta=Tc

First Law

conservation of energy; change in the internal energy of a closed system is equal to the heat and work within that system




ΔU=Q+W

+Q

heat added

-Q

heat removed

+W

work done on the gas

-W

work done by the gas

Work done on a gas

W=-PΔV

In a PV diagram, the work is equal to the

Area

Isobaric

Pressure of the gas remains constant and is a horizontal line across the graph




V/T

Isochoric (Isovolumetric)

Volume of the gas remains constant


W=0 so ΔU = Q

Isothermal

Temperature of the gas remains constant


ΔU=0 so Q=W

Adiabatic

Heat flow is zero


Q=0


ΔU=W

clockwise processes

indicate that net work is done by the gas , -W (engine)

counter-clockwise processes

indicate that net work is done on the gas, +W

Actual efficiency equation

n=Wout/Qin= 1-(Qout/Qin)

Carnot efficiency equation

Tc/Th = Th-Tc/Th

Second Law

Flow; heat naturally flows from warmer to cooler objects and the reverse process requires work from an external force




Disorder = blending


Entropy naturally increases for closed systems

Third Law

can never be cooled to absolute zero (0K)