First Law Of Thermodynamics
Heat And Thermodynamics of Class 11
Consider a system that consists of a gas enclosed by a piston in a cylinder. Suppose the system is taken quasistatically from an initial state Pi, Vi, Ti to a final state Pf, Vf, Tf. At each step the work done and heat exchanged are measured. We know that both the total work done W and the total heat transfer Q to or from the system depend on the thermodynamic path. However, the difference Q – W, is the same for all paths between the given initial and final equilibrium states, and it is equal to the change in internal energy ΔU of the system.
ΔU = Q – W (5.30)
In the above statement, Q is positive when heat enters the system and W is positive when work is done by the system on its surroundings.
The equation (5.30), is the mathematical statement of the first law of thermodynamics. It states that the internal energy of a system changes when work is done on the system (or by it), and when it exchanges heat with the environment.
Note that the first law is valid for all processes quasistatic or not. However, if friction is present, or the process is not quasistatic, the internal energy U is uniquely defined only at the initial and final equilibrium states.
The first law establishes the existence of internal energy U as a state function – one that depends only on the thermodynamic state of the system.
In the macroscopic approach of thermodynamics, there is no need to specify the physical nature of the internal energy. The experimental results are sufficient to prove that such a function exists. The internal energy is the sum of all possible kinds of energies stored in the system – mechanical, electrical, magnetic, chemical, nuclear, and so on. It does not include the kinetic and potential energies associated with the centre of mass of the system.
Misconception between Heat and Internal Energy
Confusion between heat and internal energy arises from erroneous statements that refer to the “heat content” of a body. Even correct terms like “the heat capacity of a body” can mislead one to believe that heat is somehow stored in a system. This is not correct.
The physical quantity possessed by a system is internal energy, which is the sum of all the kind of energy in the system. As the first law indicates, U may be changed either by heat exchange or by work. The internal energy is a state function that depends on the equilibrium state of a system, whereas Q and W depend on the thermodynamic path between two equilibrium states. That is, Q and W are associated with processes. The heat absorbed by a system will increase its internal energy, only some of which is average translatory kinetic energy. It is therefore incorrect to say that heat is the energy of the random motion.
APPLICATIONS OF THE FIRST LAW OF THERMODYNAMICS
We now apply the first law of thermodynamics to some simple situations.
(a) Isolated System
Consider first an isolated system for which there is no heat exchange and no work is done on the external environment. In this case Q = 0 and W = 0, so from the first law we conclude
ΔU = 0 or U = constant
The internal energy of an isolated system is constant
(b) Isochoric Process
In case of an isochoric process volume of the system remains constant
i.e.V = constant
or P/T = constant
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Since the boundary of the system does not displace because volume is constant, therefore, W = 0 The change in internal energy is given by
ΔU = nCvΔT = Using first law Q = W + ΔU
∴ Q = ΔU = |
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(5.31)
(5.32)
Here Vf = Vi
(c) Isobaric Process
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In an isobaric process pressure of The work done is given by
W = orW = Po(Vf – Vi) (5.33) Using gas equation PV = nRT We get, W =nR(Tf – Ti) |
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Since the change in internal energy is independent of the path followed, therefore ΔU = nCv ΔT = 
Using first law of thermodynamics,
Q = W + ΔU
Q = 
or Q = 
By definition, Q = nCpΔT = nCp(Tf – Ti)
Thus,Cp = 
(5.34)
Important
1.Cp – Cv = R (5.35)
2.
(5.36)
Example: 5.13
If 70 calorie of heat is required to raise the temperature of 2 mole of an ideal gas at constant pressure from 30 to 35oC, calculate
(a)the work done by the gas
(b)increase in internal energy of the gas
[R = 2 cal/mol K]
Solution
(a)At constant pressure W = P ΔV = nR ΔT (as PV = nRT)
i.e.W = nR ΔT = 2 × 2 × (35 – 30) = 20 cal
(b)ΔU = Qp - W
soΔU = 70 – 20 = 50 cal
Example: 5.14
A cylinder with a piston contains 0.2 kg of water at 100oC. What is the change in internal energy of the water when it is converted to steam at 100oC at a constant pressure of 1 atm? The density of water is ρo = 103 kg/m3 and that of steam is ρs = 0.6 kg/m3. The latent heat of vaporization of water is Lv = 2.26 × 106 J/kg.
Solution
The heat transfer to the water is
Q = mLv = (0.2 kg) (2.26 × 106 J/kg) = 4.52 × 105 J
The work done by the water when it expands against the piston at constant pressure is
W = P(Vs – Vw) = 
= (1.01 × 105 N/m2) 
= 3.36 × 104 J
The change in internal energy is
ΔU = Q – W = 452 kJ - 33.6 kJ = 418.4 kJ
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(d) Isothermal Process In an isothermal process, temperature of the system remains constant. For an ideal gas the equation of the process is given by PV = nRT = constant Work done in an isothermal process is given by
W =
or W = nRT ln |
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Since temperature of the system remains constant, therefore, there is no change in internal energy.
ΔU = nCv ΔT = 0
Using first law of thermodynamics,
Q = W + ΔU
orQ = W = nRT ln 
(5.38)
(e) Adiabatic Process
In an adiabatic process, the system does not exchange heat with the surroundings,
i.e. Q = 0.
For an ideal gas the equation of the adiabatic process is
PVγ = constant
where γ is the adiabatic exponent.
Work done: W = 
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W = Change in internal energy:
ΔU = nCvΔT = By definition, Q = 0 Therefore, using first law Q = W + ΔU ⇒0 = W + ΔU orW = -ΔU |
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(5.39)
Work done by the system is equal to the decrease in internal energy.
or-W = ΔU
Work done on the system is equal to the increase in internal energy.
Example: 5.15
Three moles of an ideal gas at 300 K are isothermally expanded to five times its volume and heated at this constant volume so that the pressure is raised to its initial value before expansion. In the whole process 83.14 kJ heat is required. Calculate the ratio (CP/CV) of the gas.
[loge5 = 1.61 and R = 8.31 J/mol K-1].
Solution
According to first law of thermodynamics,
Q = ΔU + W
For an isothermal change,
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T = constant, U = constant, ΔU = 0
and W = nRT loge i.e.W = 3 × 8.3 × 300 × loge(5) = 12.03 kJ ∴ Qisothermal = 0 + 12.03 = 12.03 kJ (i) For isochoric change as V = constant, W = ∫ P dV = 0 ΔU = nCv ΔT = 3CvΔT (n = 3) Applying gas equation between points A and C, PV/300 = P(5V) /Tc i.e.TC = 1500 K so that ΔT = TC – TB = 1500 – 300 = 1200 ∴ΔU = 3Cv × 1200 = 3.6 Cv kJ |
Note: To find TC you can apply gas equation between points B and C also. |
and hence,
(ΔQ)isochoric = 3.6 Cv + 0 = 3.6 Cv kJ (ii)
According to given problem,
ΔQisothermal + ΔQisochoric = 83.14 kJ
Using equation (i) and (ii), we get
12.03 + 3.6 Cv = 83.14
orCv = (71.11/3.6) = 19.75 J
ThusCP = CV + R = 19.75 + 8.3 = 28.05 J/mol − K
∴γ = 
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Example: 5.16
Two different adiabatic parts for the same gas intersects two isotherms at T1 and T2 as shown in the P-V diagram in the figure. How does the ratio Solution For adiabatic change PVγ = constant sincePV = nRT ∴TVγ - 1 = constant For adiabatic process, BC T1Vbγ − 1 = Τ2Vcγ - 1 (i) and for adiabatic process DA, T1Vaγ - 1 = T2Vdγ - 1 (ii) |
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compare with the ratio of 
?
Dividing equation (ii) by (i),
Hence, both ratios are same.
(f)Cyclic Process
Engines operate in cycles, in which the system – for example, a gas – periodically returns to its initial state. In figure(5.14), the system goes from state a to state b via path I, for which W1 > 0, and returns to its initial state via path II, for which WII < 0. The net work done by the system is the area enclosed by the curve.
In a clockwise traversal the net work is positive.
Since the system returns to its initial state, the change in internal energy in one complete cycle is zero, that is, ΔU = 0.
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From the first law we see that Q = W The net work done by the system in each cycle, W = WI + WII, is equal to the net heat input per cycle. This result is of importance in the discussion of steam engines and diesel engines, for instance, in which the influx of heat is used to perform mechanical work. The efficiency of a heat engine is defined as |
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(5.40)
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Example: 5.17 In the given figure, an ideal gas changes its state from A to state C by two paths ABC and AC. (i)Find the path along which work done is the least (ii)The internal energy of gas at A is 10 J and amount of heat supplied to change its state to C through the path AC is 200 J. Calculate the internal energy at C. (iii)The internal energy of gas at state B is 20 J. Find the amount of heat supplied to the gas to go from A to B. |
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Solution
Since the work done W = ∫ P dV = area under P-V curve, so
(i)WABC = WAB + WBC
i.e.WABC = 0 + 15 × 4 = 60 J
andWAC = 1/2(5 + 15) × (6 – 2) = 40 J
thus the work done along AC is least.
(ii)According to first law of thermodynamics,
dQ = dU + dW
so for path AC,
(UC – UA) = dQ – dW = 200 – 40 = 160 J
soUC = 160 + UA = 160 + 10 = 170 J
(iii)For path AB, first law of thermodynamics yields
dQ = (UB – UA) + 0 = 20 – 10 = 10 J
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Example: 5.18 One mole of oxygen undergoes a cyclic process in which volume of the gas changes 10 times within the cycle, as shown in the figure. Process : 1 - 2 and 3 - 4 are adiabatic. 2 - 3 and 4 - 1 are isochoric. Find the efficiency of the process. |
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Solution
W12 = 
Since T2 
∴T2 = T1αγ - 1 where α = V1/ V2
Thus, W12 = 
. Similarly, T3 = T4αγ - 1 and W34 = 
Also W23 = 0, W41 = 0
∴Wnet = W12 + W23 + W34 + W41 = 
Qin = 
∴η = 
Hereα = 10 ; γ = 1.4
∴η = 60%
(g) Adiabatic Free Expansion
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We now consider what happens when a gas is allowed to expand adiabatically without doing any work. Figure(5.17) shows two vessels connected by a tube with a stopcock. Initially, one vessel is filled with gas while the other is evacuated. The system is thermally insulated, that is, Q = 0. |
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When the stopcock is opened the gas quickly expands to fill the second chamber. The uncontrolled expansion is not quasistatic and cannot be depicted on a PV diagram. Since the gas does no work, W = 0. From the first law we conclude that
ΔU = 0
In an adiabatic free expansion the internal energy of any gas (ideal or real) does not change.
Example: 5.19
Two moles of an ideal monatomic gas are confined within a cylinder by a massless and frictionless spring loaded piston of cross-sectional area 4 × 10-3 m2. Initially the spring is in its relaxed state. Now the gas is heated by an electric heater, placed inside the cylinder, for some time. During this time, the gas expands and does 50 J of work in moving the piston through a distance 0.10 m. The temperature of the gas increases by 50 K. Calculate the spring constant and the heat supplied by the heater. (Po = 1.01 × 105N/m2)
Solution
(a)When the piston has displaced by x, the pressure inside the cylinder is given by
P = Po + kx/A
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i.e.50 = 105 × 4 × 10-3 × 0.1 + 1/2k × (0.1)2 or k = 2000 N/m (b)ΔU = nCv ΔT = 2 × 3/2R × 50 = 150 × 8.3 = 1245 J But from first law of thermodynamics Q = ΔU + ΔW = 1245 + 50 = 1295 J |
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Example: 5.20
A piston divides a closed gas cylinder into two parts. Initially the piston is kept pressed such that one part has a pressure Po and volume 5Vo and the other part has pressure 8Po and volume Vo; the piston is now left free. Find the new pressure and volume for the isothermal and adiabatic process (γ = 1.5).
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Solution
Final pressure will be same on both sides. Let it be P, with volume V, on the left side and Case: 1 Isothermal Process For the gas enclosed in the left chamber, Po × 5Vo = PV(i) while for the gas in the right chamber, 8Po × Vo = P(6Vo - V)(ii) After solving, we get V = 30/13Vo and P = 13/6Po and(6Vo - V) = 48/13Vo |
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Case: 2 Adiabatic Process
Po(5Vo)γ = P(V)γ(iii)
and for the gas in the right chamber,
8Po(Vo)γ = P(6Vo - V)γ (iv)
Dividing (iv) by (iii),
Substituting it in equation (iii),
P = Po
P = 1.84 Po; V = (10/3)Voand (6Vo - V) = (8/3)Vo
Example: 5.21
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A rectangular box as shown in the figure has a partition which can slide without friction along the length of the box. Initially each of the two chambers of the box has one mole of monatomic ideal gas (γ = 5/3) at a pressure Po, volume Vo and temperature To. The chamber on the left is slowly heated by an electric heater. |
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The walls of the box and the partition are thermally insulated. The gas in the left chamber expands, pushing the partition until the final pressure in both chambers becomes (243/32)Po. Calculate
(a)the final temperature of the gas in each chamber
(b)the work done by the gas in the right chamber
Solution
(a)As no heat is given to the right chamber and it is thermally insulated, so the change in the right chamber is adiabatic. And if VR is the final volume of the gas in the right chamber,
Now applying the gas equation to the gas enclosed in the right chamber
(before and after compression),
Now for the gas enclosed in the left side,
the final volume VL = 2Vo – VR = 2Vo - 
And from the gas equation
(b)Work done by the gas in the right chamber is under adiabatic condition.
WA = 
i.e.,WA = 
= -15.5 To J
Negative sign means that work is done on the gas.
Example: 5.22
One mole of an ideal monatomic gas undergoes the process p = αT1/2 , where α is a constant.
(a)Find the work done by the gas if its temperature increases by 50 K.
(b)Also, find the molar specific heat of the gas.
Solution
(a)W = ∫PdV
Using gas equation , pV = RTfor n = 1
or α√T V = RT or αV = R√T
orα dV = 
∴W = 
(use P = αT1/2)
Here,R = 8.31 J/mole/K , T2 - T1 = 50 K.
∴W =
(b)Using first law of thermodynamics
Q = ΔU + W
C(T2 - T1) = 
∴C = 
Example: 5.23
Find the maximum attainable temperature of an ideal gas in a process, P = Po - α V2, Po and α are constants, and V is the volume of one mole of gas.
Solution
Equation of the process P = P0 - αV2
Using gas equation PV = RT, we have , (P0 - αV2 ) V = RT
orT = 
For maximum value of T,
∴
