Thermodynamics lecture 7

BITS Pil iBITS PilaniPilani Campus

L t 7 W k I t l E H tLecture 7 – Work, Internal Energy, Heat

Work• From mechanics, W = ∫F.ds • δW = Fsds, generalized force and generalized sdisplacement• All forms fully inter-convertible, and ultimately reducible to the raising (or lowering) of a weight in the earth’sthe raising (or lowering) of a weight in the earth s gravitational field (not necessarily actually the case)

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Work• Work W is a form of energy transfer across the system boundary, satisfying the above criterion• W is taken as positive if the corresponding energy transfer is from the system to the surroundings, and negative if the transfer is from the surroundings to the systemtransfer is from the surroundings to the system• Unit 1J = 1 Nm• Power Ẇ = dW/dtPower Ẇ dW/dt• Unit 1W = 1 J/s • Ẇ = FV (Force x velocity) = Tω (Torque x angular velocity)


Simple Compressible SubstanceWork done at the moving boundary of a simple compressible substance

Consider a simple compressible substance at pressure P in the pistonsubstance at pressure P in the piston-cylinder arrangement as shown, undergoing a quasi-equilibrium processδW = P dV is the work done by the system in an infinitesimal change dV of the volumet e o u e

What if the process is not quasi-equilibrium, and P not defined during the process?


g pCan always use Pext, and set W = ∫δW = ∫Pext dV

Work done is path dependent

•State (point) functions and path dependent quantities (path functions) Exact and inexact differentialsfunctions). Exact and inexact differentials•Infinitesimal change in a state (point) function is an exact differential

W k d i li i l d h d d


• Work done is a line integral, and path dependent,• 1W2 = ∫ δW, with δW an inexact differential

Polytropic processA polytropic process is one in which PVn = constant C.

For such a process,

W = ∫ PdV = ∫CV-ndV

and equals

(P V P V )/(1 ) f ≠ 1(P2V2 – P1V1)/(1 – n) for n ≠ 1

P1V1 ln(V2/V1) for n = 1


Example

Ideal gas, P1 = 200 kPa, V1 = 0.04m3

(a) heated slowly at constant P to V2 = 0.1 m3

W = P1(V2 – V1) = 12 kJ1( 2 1)(b) Heated slowly, but with removal of the weights in tiny amounts such that T is constant

W = 200x0 040 ln(0 10/0 040) = 7 33 kJ


W = 200x0.040 ln(0.10/0.040) = 7.33 kJ

Example

Ideal gas, P1 = 200 kPa, V1 = 0.040 m3

(c) Process carried out in a manner that PV1.3 = constantP2 = 60.8 kPa, W = 6.41 kJ2

(d) Process carried out at constant volumeW = 0


W = 0

ExampleQuasi-equilibrium process undergone by a substance in the piston-cylinderby a substance in the piston cylinder arrangement shown. Represent the process in the P-V plane.

F↑ = F↓P A = mPg + P0 A + F1 + ks(x –x0)P = mPg/A + P0 + F1/A + ks(V –V0)/A2

= C1 + C2V

1W2 = ½ (P2 + P1)(V2 – V1)


Joule experimentFalling weight causes paddle wheel to turn in water contained in adiabatic container, and increase its t ttemperatureEstablished the mechanical equivalent of heat 1 cal = 4.182 JExperiment repeated in different variations withExperiment repeated in different variations with falling weight causing (i) Paddle wheel to turn, (ii) Two blocks to rub against one another, (iii) Compression of gas in piston-cylinder and (iv)Compression of gas in piston cylinder, and (iv) Generator to work, sending a current through a coil

In each case, the work done by the falling weight leads to the same change in state of the water. Recall that in mechanics, when the work done against a force is independent of the path that is followed, it follows


that there is a function of the position called the potential energy, the difference in which gives the work done.

Internal EnergyThe work done in the Joule experiment results in an increase in a function of state of the system (water) called its internal energy. For a closed system in an adiabatic enclosure therefore ∆U = W where W is the work done ‘by’ the system∆U = -W, where W is the work done by the systemFrom a macroscopic viewpoint, U is an extensive function of state of the system, ie., U = U(T, P), with the dimensions of y ( )energy, which is defined by the above expression for its change. It is an extension of the work-energy theorem of mechanics to now include the thermodynamic propertiesmechanics, to now include the thermodynamic properties such as T and UWhat is the microscopic interpretation of the internal energy?


Internal EnergyMicroscopically U is the sum total of all the following

contributions:contributions:

• Potential energy of interaction between the molecules

• Translational kinetic energy of the molecules

• Electronic energy of the moleculesgy

• Rotational energy of the molecules (2 degrees of freedom

for linear and 3 for non-linear molecules)for linear and 3 for non-linear molecules)

• Vibrational energy of the molecules (3n – 5 dof for linear

d 3 6 f li h i th b f t )and 3n – 6 for nonlinear where n is the number of atoms)


Heat TransferHow could one cause the same change instate of the water in the Joule experiment without performance of work?• By making one of the walls diathermal, and placing in thermal contact with another system at a higher temperature

Si k f d W 0 H i U i t t• Since no work performed, W = 0. However, since U is a state function, and the same change in state has been caused, necessarily ∆U ≠ 0, and has the same value as before. How yhas this change come about? • By a transfer of heat Q = ∆U from the system at a higher temperature Heat is a mode of energy transfer that occurstemperature. Heat is a mode of energy transfer that occurs when two bodies at different temperatures are in thermal contact.


Heat• Heat Q is energy transfer that occurs solely due to a

temperature difference between two systemstemperature difference between two systems

• Q has units of energy, J (calorie also used traditionally)

• Q taken as positive when energy transfer to system results

• Energy transfer as heat Q from system at higher temperature gy y g p

to system at lower temperature

• Q = ∫δQ is path dependent and δQ is inexact• Q = ∫δQ is path dependent, and δQ is inexact

• q = Q/m is heat transfer per unit mass

• Can the Joule experiment be reversed entirely?


First Law for Control MassAs seen above, for a closed system

∆U U U W ( di b ti Q 0)∆U = U2 – U1 = - 1W2 (adiabatic, Q = 0)

∆U = U2 – U1 = 1Q2 (W = 0)

What if both work and heat terms present? Since U is a state

function

∆U = U2 – U1 = 1Q2 - 1W2

More generally since there may also changes in kinetic energy of

the system, and its potential energy in the gravitational field, we

use the total energy E = U + KE + PE, and write

E2 – E1 = 1Q2 - 1W2


Work and Heat• Both heat and work are transient phenomena, not

properties

• Both are modes of energy transfer across system

boundary when system undergoes a process

• Both are path dependent and the infinitesimals δQ andBoth are path dependent, and the infinitesimals δQ and

δW are inexact differentials


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Thermodynamics lecture 7