# Physics Subject Area Test Thermodynamics. There are three commonly used temperature scales, Fahrenheit, Celsius and Kelvin.

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Physics Subject Area Test Thermodynamics

There are three commonly used temperature scales, Fahrenheit, Celsius and Kelvin.

Converting Between the Kelvin and Celsius Scales Converting Between the Fahrenheit and Celsius Scales

Thermal expansion Expanding solids maintain original shape Expanding liquids conform to the container Linear expansion ΔL = αLΔTL = length α = coefficient of liner expansion ΔT = temperature change

Example: The highest tower in the world is the steel radio mast of Warsaw Radio in Poland, which has a height if 646m. How much does its height increase between a cold winter day when the temperature is -35⁰C and a hot summer day when the temperature is +35 ⁰C ? ΔL = αLΔT = 12x10 -6 / ⁰C x 646 x 70 ⁰C = 0.54m

Volume expansion ΔV= βVΔTL = length β = coefficient of liner expansion ΔT = temperature change coldhot Β = 3 α

Heat flow: the heat current; the amount of heat that passes by some given place on the rod per unit time

* convection Heat is stored in a moving fluid and is carried from one place to another by the motion of this fluid * radiation The heat is carried from one place to another by electromagnetic waves * conduction the process of handing on energy from one thing to the next

P 1 V 1 =P 2 V 2 P= pressure V = volume

V 1 /V 2 =T 1 /T 2 V = volume T = temperature

P 1 /T 1 =P 2 /T 2 P = pressure T = temperature

Ideal Gas Law PV = n R T P= pressure V = volume T = temperature n = moles R = Gas constant = 0.08206 L-atm/mol K

PV = nRT n/V = P/RT Molarity = n/V Density D = m/V Molecular Wt M = m/n D=Mn/V = PM/RT M= DRT/P

Energy can be neither created nor destroyed but only transformed

THE GENERAL ENERGY EQUATION Energy In = Energy Out or U 2 - U 1 = Q -W where U 1 : internal energy of the system at the beginning U 2 : internal energy of the system at the end Q : net heat flow into the system W : net work done by the system Q = ΔU + ΔW

A closed tank has a volume of 40.0 m 2 and is filled with air at 25 ⁰C and 100 kPa. We want to maintain the temperature in the tank at 25 ⁰C as water is pumped into it. How much heat will have to be removed from the air in the tank to fill it half full? = (100kPa) (40.0 m 2 )(-0.69314) = -2772.58kJ

Isobaric –the pressure of and on the working fluid is constant –represented by horizontal lines on a graph Isothermal –temperature is constant –Temperature doesn’t change, internal energy remains constant, & the heat absorbed by the gas = the work done by the gas –The PV curve is a hyperbola Adiabatic –there is no transfer of heat to or from the system during the process –Work done = decrease in internal energy & the temperature falls as the gas expands –-the PV curve is steeper than that of and isothermal expansion

Quasi-static (quasi-equilibrium) processes – sufficiently slow processes, any intermediate state can be considered as an equilibrium state (the macroparamers are well- defined for all intermediate states). Examples of quasi- equilibrium processes: isochoric: V = const isobaric: P = const isothermal: T = const adiabatic: Q = 0 For quasi-equilibrium processes, P, V, T are well-defined – the “path” between two states is a continuous lines in the P, V, T space. P V T 1 2 Advantage: the state of a system that participates in a quasi-equilibrium process can be described with the same (small) number of macro parameters as for a system in equilibrium (e.g., for an ideal gas in quasi- equilibrium processes, this could be T and P). By contrast, for non- equilibrium processes (e.g. turbulent flow of gas), we need a huge number of macro parameters.

The sign: if the volume is decreased, W is positive (by compressing gas, we increase its internal energy); if the volume is increased, W is negative (the gas decreases its internal energy by doing some work on the environment). The work done by an external force on a gas enclosed within a cylinder fitted with a piston: W = (PA) dx = P (Adx) = - PdV xx P W = - PdV - applies to any shape of system boundary The work is not necessarily associated with the volume changes – e.g., in the Joule’s experiments on determining the “mechanical equivalent of heat”, the system (water) was heated by stirring. dU = Q – PdV A – the piston area force

* Specific Heat the heat absorbed during the change of state Q = nC v ΔTQ = amount of heat required n = number of moles C v = specific heat at a constant volume ΔT = Change in temperature

How to calculate changes in thermal energy Specific heat is the amount of heat required to raise the temperature of 1 kg of a material by one degree (C or K). C water = 4184 J / kg C Q = m x  T x C p Q = change in thermal energy m = mass of substance  T = change in temperature (T f – T i ) C p = specific heat of substance

Second Law of Thermodynamics Entropy = the transformation of energy to a more disordered state - can be thought of as a measure of the randomness of a system - related to the various modes of motion in molecules The second law of thermodynamics: entropy of an isolated system not in equilibrium tends to increase over time No machine is 100% efficient Heat cannot spontaneously pass from a colder to a hotter object

The relationship between kinetic energy and intermolecular forces determines whether a collection of molecules will be a solid, liquid or a gas * Pressure results from collisions * The # of collisions and the KE contribute to pressure * Temperature increase KE

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