6.1. FIRST LAW FOR A CYCLIC PROCESS OF A CLOSED SYSTEM
Between 1843 and 1848, James P. Joule carried out experiments which were the first step in the analysis of thermodynamics system leading to the discovery of the First Law of Thermodynamics. This law is the law of conservation of energy i.e. energy can neither be created nor destroyed but only converted from one form to another.
One such experiment is given below:
6.1.1. Joule's experiment:
Step-I: A known mass of fluid in an inner vessel is placed inside the water ice bath (at 0°C temperature) as shown in Fig. 6.1 (a). Consider fluid in the inner vessel as a system. Note down the temperature of fluid of inner vessel when the fluid of inner vessel comes in thermal equilibrium with water bath.
Fig. 6.1 (a)
Step-II: Lift the inner vessel outside the water bath and place it in some insulation. Now, work is done on the fluid by stirring the water with paddle wheel by a falling weight from height ‘A’ to ‘B’ as shown in Fig. 6.1 (b). The work input to the fluid causes a rise in the temperature of the fluid from 0°C. The amount of work added in the system is calculated by the product of weight and the vertical height through which the weight descends.
Fig. 6.1 (b)
Step-III: Next, in order to restore the system to its original state, the inner vessel, by removing its insulation, is again immersed in the water ice bath as shown in Fig. 6.1 (c). While in contact with the water bath, the heat is transferred from fluid to the water ice bath as the temperature of fluid is higher than water ice bath and heat transfer takes place till the fluid and water ice bath come in thermal equilibrium i.e. the original state of fluid is reached. This way the system has undergone a cyclic process. The amount of heat flow out of the fluid is measured by the product of mass of ice melted in the water ice bath and latent heat of ice.
Fig. 6.1 (c)
6.1.2. Joule's experiment conclusion:
This experiment lead to an important conclusion that the net quantity of work done on the system was directly proportional to the net quantity of heat removed from the system during the cycle.
Joule conducted many similar experiments with different systems involving different work interactions and found the same results.
Thus for a closed system undergoing cyclic process
If both heat and work are expressed in the same unit of J or kJ, then it becomes
From this observation Joule postulated the First Law of Thermodynamics which states that if a closed system undergoes a cyclic process, energies transacted by it in the form of work and heat are equal.
Mathematically, 
(Units: J or kJ)
In a cycle involving two processes, the First Law is expressed as
Q1-2 + Q2-1 = W1-2 + W2-1
For a four process cycle, it is expressed as
Q1-2 + Q2-3 + Q3-4 + Q4-1 = W1-2 + W2-3 + W3-4 + W4-1
6.1.3. Corollary of First Law: It is impossible to construct a perpetual motion machine of first kind (PMM-1). A Perpetual Motion Machine of the First kind (PMM-1) PMM-1 is capable of producing energy without corresponding expenditure of energy. The converse of this is also true. Example: Refer Fig 6.2. The overbalancing wheel mechanism once sets in motion will continue to run due to the weighted arms with metal balls falling in one direction of the turning wheel. This is Perpetual Motion Machine of the First kind. | Fig. 6.2. PPM -1 |
6.2. FIRST LAW OF THERMODYNAMICS UNDERGOING A PROCESS i.e. CHANGE OF STATE OF A CLOSED SYSTEM
Consider a system in state ‘1’. Now by flowing energy in the form of work and heat in or out of a system, it undergoes a change of state to state ‘2’ through path ‘A’ (Fig. 6.3). Now by again transferring work and heat in or out of a system, it has the option to come back to its initial state ‘1’ through two paths ‘B’ and ‘C’ thus to complete a cycle. |
Fig. 6.3. Existence of the thermodynamics property E. |
For a thermodynamic cycle through paths A and B, according to first law of thermodynamics
.......................................... (6.1)
And for a thermodynamic cycle through paths A and C
.................................................... (6.2)
Subtract equation (6.2) from equation (6.1), we have
Since B and C are two different paths between states 1 and 2 and quantity (δQ - δW) is the same for both these paths. Therefore, (δQ-δW) does not depend on the path, it depends only on states. We conclude that (δQ-δW) is a point function, and therefore is the differential of a property of a system. This property is the total energy of the system and is given the symbol E.
Thus, (δQ – δW) = dE
or δQ = dE + δW
For process 1-2, the above equation is integrated from state 1 to state 2,
or 1Q2 = (E2 – E1) + 1W2 …………………………..(6.3)
where, 1Q2 is the heat transfer to the system during the process from state 1 to state 2
E2 and E1 is the total energy of the system in state 1 and state 2, respectively.
1W2 is the work done by the system during the process from state 1 to state 2
The total energy of the system (E)
E = K.E. + P.E. + U
Where, KE is the Kinetic Energy possessed by a system as a whole with respect to an external reference frame (Macroscopic form) = ½ mV.
PE is the Potential Energy possessed by a system as a whole with respect to an external reference frame (Macroscopic form) = mgZ.
U (internal energy) is the sum of kinetic and potential energies possessed by the atoms and molecules present in the gas (Microscopic form).
or dE = dU + d(K.E.) + d(P.E.)
For process 1-2, the above equation can be integrated from state 1 to state 2 as,
(E2 – E1) = (U2 – U1) + ½ m(V22 – V12)+ mg(Z2 – Z1) , (J or kJ)
By substituting the value of (E2 – E1) in equation (6.3), we get
1Q2 = (U2 – U1) + ½ m(V22 – V12)+ mg(Z2 – Z1) + 1W2 ...............………………..(6.4)
or (U2 – U1) + ½ m(V22 – V12)+ mg(Z2 – Z1) = 1Q2 - 1W2
The above equation states that when the energy in the form of work and heat go in or out of a system by crossing the system boundary, the net change in energy of the system (E) for a change of state will be exactly equal to the net energy that crossed the system boundary in the form of work and heat. The net energy of a system may change in internal energy, kinetic energy, or potential energy.
In the absence of kinetic energy and potential energy, ½ m(V22 – V12)= 0 and mg(Z2 – Z1) = 0,
Equation (6.4) is simplified to 1Q2 = (U2 – U1) + 1W2 …………………. (6.5)
The above equation (6.5) can be written per unit basis as
1q2 = (u 2 – u 1) + 1w2 (J/kg or kJ/kg)
where 1w2 = 
+ (other ways of work done by a system)
6.3. INTERNAL ENERGY OF AN ISOLATED SYSTEM
As we know an isolated system is one in which there is no interaction of the system with the surroundings either through heat or work modes, we can write
i.e. Q = 0 and W = 0
From equation (6.5),
(U2 – U1) = 0
or U2 = U1
6.4. INTERNAL ENERGY OF A PERFECT GAS
Joule proved experimentally that the internal energy of a perfect gas is a function of absolute temperature alone and it is independent of change in pressure and volume,
i.e. U = f(T) only.
dU = m CvdT
where, m = mass of a gas, kg
dU = change in specific internal energy of a gas, kJ/kg
dT = change in absolute temperature of a gas between two states, K
Cv = Specific heat of a gas at constant volume, J/kg K or kJ/kg K
For a process 1-2, the above equation is integrated from state 1 to state 2,
U2 – U1 = m Cv (T2 – T1)
On per unit mass basis, the equation (6.5) can be written as
u2 – u1 = Cv (T2 – T1)
Problem 6.1: A tank containing a fluid is stirred by a paddle wheel. The power input to the paddle wheel is 1500 W. Heat is transferred from the tank at the rate of 450 W. Considering the tank and the fluid as the system, determine the change in the internal energy of the system during 1 sec.
Solution:
Given: lW2 = - 1500 W; 1Q2= - 450 W
To determine change in the internal energy of the system during 1 sec:
Formula: Apply the first law of thermodynamics for a change in state
1Q2 = (U2 – U1 ) + 1W2
or (U2 – U1 ) = 1Q2 – 1W2
Answer: The change in the internal energy of the system during 1 sec
(U2 – U1 ) = 1Q2 – W2
(U2 – U1) = - 450 – (–1500)
(U2 – U1) = 1050 W
or U2 – U1 (in one sec.) = 1050 J
Problem 6.2: A system undergoes a cycle that comprises the four processes 1-2, 2-3, 3-4 and 4-1. The energy transfers are tabulated below. Determine the unknown kJ and complete the table.
Process i-f | iQf (kJ) | iWf (kJ) | ΔU =Uf - Ui (kJ) |
1-2 | 40 | --- | 25 |
2-3 | 20 | – 10 | --- |
3-4 | – 20 | ---- | --- |
4-1 | 0 | 8 | --- |
where, f- final and i – initial
Solution:
Given: 1Q2= 40 kJ; (U2 – U1) = 25 kJ; 2Q3= 20 kJ; 2W3= - 10 kJ; 3Q4= - 20kJ; 4Q1= 0 kJ; 4W1= 8 kJ.
(a) Process 1-2:
To determine work, 1W2
Formula: Apply the first law of thermodynamics.
1Q2 = (U2 – U1) + ½ m(V22 – V12)+ mg(Z2 – Z1) + 1W2
or1 W2 = 1Q2 – (U2 – U1)
Answer: 1W2 = 1Q2 – (U2 – U1)= 40 – 25 = 15 kJ
(b) Process 2-3:
To determine work, U3 – U2
Formula: Apply the first law of thermodynamics.
2Q3 = (U3 – U2) + ½ m(V32 – V22)+ mg(Z3 – Z2) + 2W3
or (U3 – U2) = 2Q3 – 2W3
Answer: (U3 – U2 ) = 2Q3 - 2W3 =20 -(-10) = 30 kJ
(c) Process 4-1:
To determine work, U1 – U4
Formula: Apply the first law of thermodynamics.
4Q1 = (U1 – U4) + ½ m(V12 – V42)+ mg(Z1 – Z4) + 4W1
(U1 – U4) = 4Q1 – 4W1
Answer: (U1 – U4 ) = 4Q1 - 4W1 =0 – 8 = – 8 kJ
(d ) Process 3-4:
To determine work, U4 – U3
Formula: As cyclic 
;
(U2 – U1) + (U3 – U2)+ (U4 – U3) + (U1 – U4) = 0
or U4 – U3 = 0 – [(U2 – U1) + (U3 – U2) + (U1 – U4)]
Answer: U4 – U3 = 0 – [(U2 – U1) + (U3 – U2) + (U1 – U4)]
= 0 – [25 + 30+(-8)] = – 47 kJ
To determine work, 3W4
Formula: Apply the first law of thermodynamics.
3Q4 = (U4 – U3) + ½ m(V42 – V32)+ mg(Z4 – Z3) + 3W4
or 3W4 = 3Q4 – (U4 – U3)
Answer: 3W4 = 3Q4 – (U4 – U3) = - 20 – (–47) = 27 kJ
Answer: The complete table is:
Process | Q (kJ) | W (kJ) | ΔU = Uf - Ui (kJ) |
1-2 | 40 | 15 | 25 |
2-3 | 20 | - 10 | 30 |
3-4 | - 20 | 27 | - 47 |
4-1 | 0 | 8 | - 8 |
Problem 6.3. Considering system composed of a stone having a mass of 10 kg and a bucket containing 100kg of water. Initially the stone is 10 m above the water and the stone and water are at the same temperature. The stone then falls into the water.
Determine ΔU, ΔKE, ΔPE, δQ and δW for the cases:
(a) At the instant the stone is about to enter the water. Assuming no heat transfer to or from the stone as it falls
(b) Just after the stone come to the rest in the bucket
(c) After enough heat has been transferred so that the stone and water are at the same temperature they were initially.
Solution: Consider stone and water as a system.
Given: Mass of stone = 10 kg; Mass of water = 10 kg
(a) Change of state from stone 10 m above the water to stone is about to enter the water
Given: 1 Q2 = 0; ΔU=0 ; 1W2 =0;
Initial position of stone = 10 m above water
Determine change in kinetic energy ΔKE and potential energy ΔPE:
Formula: Apply the first law of thermodynamics for the closed systems.
1Q2 = (U2 – U1) + ½ m(V22 – V12)+ mg(Z2 – Z1) + 1W2
The above equation reduces to
− ΔKE = ΔPE = mg(Z2 – Z1)
Answer: − ΔKE = ΔPE = mg(Z2 – Z1) =10 x 9.8 x (−10) = − 980 J
(b) Change of state from stone 10 m above the water to stone come to the rest in the bucket:
Given: 1Q2 = 0; 1W2 = 0; ΔKE = 0; Determine change in potential energy ΔPE and internal energyΔU: Formula: Apply the first law of thermodynamics for the closed systems. 1Q2 = (U2 – U1) + ½ m(V22 – V12) + mg(Z2 – Z1) + 1W2 The above equation reduces to − ΔU = ΔPE = mg(Z2 – Z1) Answer: - ΔU = ΔPE = mg(Z2 – Z1) =10 × 9.8 × (−10) = −980 J or ΔU = 980 J; ΔPE = − 980 J |
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(c) Change of state from stone 10 m above the water to water are at the same temperature it was initially due to heat transfer:
Given: ΔU = 0; ΔKE = 0 ; 1W2 =0;
Determine heat transfer 1Q2:
Formula: Apply the first law of thermodynamics for the closed systems.
1Q2 = (U2 – U1) + ½ m(V22 – V12)+ mg(Z2 – Z1) + 1W2
The above equation reduces to
1Q2 = mg(Z2 – Z1)
Answer: 1Q2 = mg(Z2 – Z1) =10 * 9.8 * (−10) = − 980 J
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