2. Work and heat oWhen 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). oWork 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. oWork 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.

3. Quasi-static processes oThe 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. oWhen 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. oThe 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.

4. 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 V i to V f : f i

5. Work done is path dependent The 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.

6. 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, T i ) to the same final state (2V, P/2, T i ) : 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.

7. 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.

8. When a system undergoes a change from one state to another, the change in its internal energy is:  E int = E f – E i = 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

9. 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.  E int is path-independent and the internal energy E is a state property. For an isolated system (one that does not interact with its surroundings),  E int = 0 since there is neither heat transfer nor work done (i.e. Q = W = 0)

10. 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

11. 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:  E int = – 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.

12. 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:  E int = Q This means that the heat that flows into the system has served to increase its internal energy.

13. 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 (V f – V i )

14. 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.

15. 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 V i to the final volume V f ? The work done is: f i For an ideal gas: PV = nR T. Hence, f f i i i f

16. 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? 86428642 0 2 4 6 8 10 C B A V(m 3 ) P(kPa) In a cyclic process  E int = Q – W = 0 ThereforeQ = W = area of the triangle ABC = (6 kPa)·(4 m 3 )/2 = 12 kNm = 12 kJ

17. In the earlier example, if Q is negative for the process BC and  E int 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   E int = Q – W < 0   E int < 0 for both process BC and process CA   E int > 0 for process AB because  E int = 0 for the whole cycle process  For process AB:  E int > 0 and W > 0  Q > 0 (note: Q =  E int + W)

18. 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 P o and V o, the net energy transferred by heat to the gas in this cycle. A C D B V 3Po3Po PoPo P VoVo 2V o T1T1 T2T2 A  B: W 1 = P A (V B – V A ) B  C: W 2 = nRT 2 ln(V C /V B ) C  D: W 3 = P C (V D – V C ) D  A: W 4 = nRT 1 ln(V A /V D ) Since P A V B = P C V C and P A V A = P C V D W =  W i = W 2 + W 4 = nRT 2 ln(V C /V B ) + nRT 1 ln(V A /V D ) W = nRT 2 ln(V C /V B ) - nRT 1 ln(V D /V A ) = nRT 2 ln(P A /P C ) - nRT 1 ln(P A /P C ) = P C V C ln(P A /P C ) - P D V D ln(P A /P C ) = 2P o V o ln(3) - P o V o ln(3) = P o V o ln(3)