Download - Phys 160 Thermodynamics and Statistical Physics · Heat and Work Much of thermodynamics deals with three closely -related concepts; temperature, energy, and heat. •Temperature,

Phys 160 Thermodynamics and Statistical

Physics

Lecture 4 Isothermal and Adiabatic Work

Heat Capacities

Heat and Work Much of thermodynamics deals with three closely -related concepts; temperature, energy, and heat.

• Temperature, fundamentally, is a measure of an object's tendency to spontaneously give up energy. • Energy is the most fundamental

dynamical concept in all of physics

• For this reason, how to express energy in terms of anything more fundamental?

• We can only state what we know definitely. The law of conservation of energy.

• There are many ways to put energy into a system or taken out.

• In Thermodynamics, we classify these mechanisms as Heat and Work

• Heat is defined as spontaneous flow of energy from one object to another due to difference in temperature. • Work is defined as any other

transfer of energy into or out of a system.

• Usually with work an agent is involved in putting in energy; it is not spontaneous

• Both heat and work refers to energy in transit; there is no heat or work in a system. Heat enters a system; work is done on a system. .

• Define symbols: • U for the total energy inside a

system.

• Q and W represent the energy that enters and leave the system as heat and work. Q + W is the total energy that enters the system

• By the conservation law, • U = Q + W . U is the change in energy.

• Can we say similarly change in Q and W? Q and W are infinitesimal quantities.

• This statement U = Q + W is what is know as the first law of thermodynamics

• The SI unit for energy is the Joule kg.m2/s2. Traditional unit is calorie. 1 cal = 4.186 J.

• Compression Work –important type of work done on a system (gas)

• W = F. dr = F x =P.A. x (volume change should be very slow-quasistatic)

• Why negative sign in front of PdV?

• The work done here is positive but the dV is negative, compression

• Hence W = -PdV

• isothermal

• For the isothermal (limits Vi to Vf) • W = - PdV =-NkT V/V

• = NkT ln (Vi/Vf)

• The work done here is positive.

• Because it is isothermal, heat must flow out of the system. How much? How to calculate?

• From energy conservation, Q = U – W = (NfkT/2) – W There is no change in U; hence U = 0 and Q = 0 – W = NkTln(Vf/Vi) • The heat input here is just the

negative of work done. expansion

• For isothermal compression, Q is nega-tive since heat leaves the gas. For iso-thermal expansion, Q is positive since heat enters the system. • Adiabatic Compression. We will assume

quasistatic process, but no heat is allow-ed to escape. For this U = Q + W =W;

• T will increase. • The curve in the PV diagram must con-

nect a lower T isotherm with one of higher T, its slope higher than either.

• What is the shape of this curve?

• From equipartition theorem • U =(f/2) NkT. Then the energy change

along any infinitesimal segment of the curve is dU = (f/2)N k dT = - P dV

• This differential equation relates dT to dV. To solve the equation use the ideal gas equation P = NkT/V to eliminate P

• The solution to this equation is (f/2) (dT/T) = - ( dV/V) Integrating (f/2) ln (Tf/Ti) = -ln (Vf /Vi)

• Exponentiating we have • Vf Tf

f/2 = ViTi f/2 = V T

f/2 • Squaring we have V2 T

f = constant. Eliminating T using PV = T* constant, we have V2 (PV)f =constant. Taking the fth root, P V (f+2)/f = constant = PV where is called the adiabatic exponent.

• An ideal gas is made to undergo the cyclic process as in Fig. For each step A, B and C: (a) Find the work done on the gas (b) the change in the energy content of the gas (c) the heat added to the gas. What does this process accomplish.