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

• 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.

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

• 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.