Work and heato When an object is heated and its volume is allowed

to expand, then work is done by the object and the amount of work done depends generally on the pressure (P) and change of volume (V).

o Work done by an object or a thermodynamic system is commonly taken to be positive, and work done on an object or a thermodynamic system is conventionally taken to be negative.

o Work is just another form of energy. Hence, the work done by a system represents a transfer of energy out of the system, and the work done on a system represents a transfer of energy to the system.

Quasi-static processeso The thermodynamic properties (such as temperature,

pressure, volume and internal energy) of a system can be specified only if the system is in thermal equilibrium internally, namely the thermodynamic properties of every part of the system are the same.

o When a system is brought from one equilibrium state to another equilibrium state through a thermodynamic process, its thermodynamic properties change during the process and the system may not be in thermal equilibrium internally at times.

o The word quasi-static is used to describe processes that are carried out slowly enough so that the system passes through a continuous sequence of thermal equilibrium states.

Work done by quasi-static expansion of a gas

Work done by the expanding gas in moving the piston up by dy:

dW = F dy = PA dy (A = cross-sectional area of piston) = P dV

Total work done by the gas as its volume changes from Vi to Vf :

V

VPdVW f

i

Work done is path dependentThe work done in bring the gas from an initial state i to a final state f depends on the path taken between the two states, as illustrated in the following PV diagrams of three different processes:

It is obvious that the work done also depends on the initial and final states.

Energy transferred by heat in bring a system from an initial state i to a final state f is also process-dependent (or path-dependent). This can be easily illustrated with the following two processes that bring a gas from an initial state (V, P, Ti) to the same final state (2V, P/2, Ti) :

The force on the piston is reduced so slowly as to maintain a constant temperature. During the process, heat flows into gas and the gas does work.

The membrane is broken to allow rapid expansion of the gas into the vacuum. No work is done by the gas and there is no heat transfer during the process.

While both the heat transferred and the work done in bring a system from one state to another is path-dependent, the quantity “Q – W” is not. Experimental evidences show that “Q – W” depends only on the initial and final states on the system.

Hence for a cyclic process (a process that originates and terminates at the same state), Q – W = 0 or Q = W.

When a system undergoes a change from one state to another, the change in its internal energy is:

Eint = Ef – Ei = Q – W

where Q is the energy transferred into the system and W is the work done by the system.

The first law of thermodynamics

• The first law of thermodynamic is essentially the law of energy conservation that includes changes in internal energy. In other words, the net energy gained (Q – W) by the system during a process is converted into internal energy.

Eint is path-independent and the internal energy E is a state property.

• For an isolated system (one that does not interact with its surroundings), Eint = 0 since there is neither heat transfer nor work done (i.e. Q = W = 0)

When a system undergoes only an infinitestimal change in state, only infinitestimal amount of heat is absorbed and only an infinitestimal amount of work is done, so that the internal energy change is also infinitestimal. In such a case, the first law is written in differential form as:

dE = dQ – dW

In systems where the work is given by PdV, the first law in differential form becomes:

dE = dQ - PdV

Adiabatic process

In an adiabatic process, there is no heat transfer between the system and its surroundings (i.e. Q = 0). In such a process, the first law gives:

Eint = – W

The internal energy increases if work is done to the system, and this usually leads to a rise in temperature. If work is done by the system, the internal energy decreases which is usually accompanied by a temperature drop.

Isochoric (isovolumetric) process

In an isochoric process, the volume of the system remains unchanged (i.e. V = 0), implying no work is done. In such a process, the first law gives:

Eint = Q

This means that the heat that flows into the system has served to increase its internal energy.

Isobaric process

In an isobaric process, the pressure of the system is constant. In such a process, the work done by the system is:

W = P (Vf – Vi)

Isothermal process

In an isothermal process, the temperature of the system remains the same all the time. In order for the temperature to remain constant, the changes in the pressure and the volume must be kept very slow so that the process is quasi-static. In general, none of the quantities Q, W and E is zero.

Example : What is the work done by an ideal gas when it is allowed to expand quasi-statically at constant temperature from the initial volume Vi to the final volume Vf ?

The work done is: V

VPdVW f

i

For an ideal gas: PV = nR T.Hence,

V

VnRTdV

VnRTdV

V

nRTW V

V

V

Vln1 ff

iii

f

Example : A gas is taken through the cyclic process as shown in the figure below. What is the heat energy transferred to the system in one complete cycle?

8

6

4

2

0 2 4 6 8 10

C

B

A

V(m3)

P(kPa)

In a cyclic process Eint = Q – W = 0

Therefore Q = W = area of the triangle ABC = (6 kPa)·(4 m3)/2

= 12 kNm = 12 kJ

In the earlier example, if Q is negative for the process BC and Eint is also negative for the process CA, What is the sign of Q for the process AB?

For process BC: Q < 0 and W = 0 Eint = Q – W < 0

Eint < 0 for both process BC and process CA Eint > 0 for process AB because Eint = 0 for the whole cycle process

For process AB: Eint > 0 and W > 0 Q > 0 (note: Q = Eint + W)

Example : An ideal gas is carried through a thermodynamic cycle ABCD consisting of two isobaric and two isothermal processes as shown in the figure below. Determine, in terms of Po and Vo, the net energy transferred by heat to the gas in this cycle.

A

CD

B

V

3Po

Po

P

Vo 2Vo

T1

T2

A B: W1 = PA(VB – VA)

B C: W2 = nRT2 ln(VC /VB)

C D: W3 = PC(VD – VC)

D A: W4 = nRT1 ln(VA /VD)

Since PAVB = PCVC and PAVA = PCVD

W = Wi = W2 + W4 = nRT2 ln(VC /VB) + nRT1 ln(VA /VD)

W = nRT2 ln(VC /VB) - nRT1 ln(VD /VA) = nRT2 ln(PA /PC) - nRT1 ln(PA /PC) = PCVC ln(PA /PC) - PDVD ln(PA /PC) = 2PoVo ln(3) - PoVo ln(3) = PoVo ln(3)