Thermal Engineering - I - Unit 1 Thermodynamics Process of Perfect Gases - N Scheme III Sem Diploma Mech

6 April 2022 U.Aravind, Lect/Mech, LAPC 1
Lakshmi Ammal Polytechnic College
Department of Mechanical Engineering
Subject Code : 4020340
Year & Semester : II / III
Scheme : N
By…
U.Aravind,
Lecturer/Mechanical Engineering,
LAPC, Kovilpatti.

Lakshmi Ammal Polytechnic College
Department of Mechanical Engineering
Subject Code : 4020340
Subject Name : Thermal Engineering I
Year & Semester : II / III
Scheme : N
Unit : 1
Topic: II. Thermodynamics Processes of
Perfect Gases

Thermodynamic Processes
The Study of thermodynamic process
involves the determination of the
Work (W)
Change in Internal energy (∆U)
Change in Entropy (∆S)
The amount of heat transferred (Q)

Thermodynamic Processes
•Constant Volume (or) Isochoric Process [V=c]
•Constant Pressure (or) Isobaric Process [p=c]
•Constant Temperature (or) Isothermal (or)
Hyperbolic Process [T=c]
•Reversible adiabatic (or) Isentropic Process
[pvγ
= c]
•Polytropic Process [pv𝑛
= c]
•Free Expansion Process
•Throttling Process

• m kg of gas taken in a Closed Vessel.
• It is heated or cooled
• Being contained in a closed vessel gas neither
expands nor compressed.
• So volume remains constant.

i) General gas equation :
𝐏𝟏𝐕𝟏
𝐓𝟏
=
𝐏𝟐𝐕𝟐
𝐓𝟐
Apply 𝐕𝟏 = 𝐕𝟐 in to General Gas Equation
𝐏𝟏𝐕𝟏
𝐓𝟏
=
𝐏𝟐𝐕𝟐
𝐓𝟐
𝐏𝟏
𝐓𝟏
=
𝐏𝟐
𝐓𝟐
ii) Heat Transferred(Q)
Heat Transferred = mass x specific heat at constant volume
x temperature difference
Q = m 𝐂𝐯 (T2 – T1)
iii) Work done (W)
As gas neither expands nor compressed in a closed vessel, so
work done is Zero.
W = 0

iv) Change in Internal Energy (∆U)
Change in Internal Energy= mass x specific heat at constant volume
x temperature difference
∆U = m 𝐂𝐯 (T2 – T1) (Common for all process)
v) By first law of thermodynamics
Q = ∆U + W
= ∆U + 0
Q= ∆U
vi) Change in Entropy (∆S)
∆S = 𝐓𝟏
𝐓𝟐 𝐝𝐐
𝐓
Q = m𝐂𝐯 (T2 – T1)
In differentiate form
dQ = m𝐂𝐯 dT
= 𝐓𝟏
𝐓𝟐 m𝐂𝐯 dT
𝐓
𝟏
𝐱
dx = 𝐥𝐨𝐠𝒆𝒙 = 𝐥𝐧𝐱
∆S = m𝐂𝐯 ln
𝑻𝟐
𝑻𝟏

• m kg of gas taken in an engine cylinder.
• To maintain constant pressure some weight is
placed on the piston.
• Gas is heated
• It expands and does work on the pressure.

𝐏𝟏𝐕𝟏
𝐓𝟏
=
𝐏𝟐𝐕𝟐
𝐓𝟐
Apply 𝐏𝟏 = 𝐏𝟐 in to General Gas Equation
𝐏𝟏𝐕𝟏
𝐓𝟏
=
𝐏𝟐𝐕𝟐
𝐓𝟐
𝐕𝟏
𝐓𝟏
=
𝐕𝟐
𝐓𝟐
Heat Transferred = mass x specific heat at constant
pressure x temperature difference
Q = m 𝐂𝐏 (T2 – T1)
iii) Work done (W)
Work done = hatched area in PV diagram = height x breadth
W = 𝐏𝟏(𝐕𝟐 - 𝐕𝟏)

iv) By first law of thermodynamics
Q= ∆U + W
v) Change in Entropy (∆S)
∆S = 𝐓𝟏
𝐓𝟐 𝐝𝐐
𝐓
Q = m𝐂𝐏 (T2 – T1)
In differentiate form
dQ = m𝐂𝐏 dT
= 𝐓𝟏
𝐓𝟐 m𝐂𝐏 dT
𝐓
𝟏
𝐱
dx = 𝐥𝐨𝐠𝒆𝒙 = 𝐥𝐧𝐱
∆S = m𝐂𝐏 ln
𝑻𝟐
𝑻𝟏

• It is heated slowly by source
• Temperature may tend to raise. But this is
nullified by the expansion of gas.
• So the temperature remains constant.

𝐏𝟏𝐕𝟏
𝐓𝟏
=
𝐏𝟐𝐕𝟐
𝐓𝟐
Apply 𝐓𝟏 = 𝐓𝟐 in to General Gas Equation
𝐏𝟏𝐕𝟏
𝐓𝟏
=
𝐏𝟐𝐕𝟐
𝐓𝟐
𝐏𝟏𝐕𝟏 = 𝐏𝟐𝐕𝟐
ii) Change in Internal Energy (∆U)
∆U = m 𝐂𝐯 (T2 – T1)
𝐓𝟏 = 𝐓𝟐
∆U = 0
iii) Change in Enthalpy (∆h)
∆h = m 𝐂𝐏 (T2 – T1)
𝐓𝟏 = 𝐓𝟐
∆h = 0
iv) By first law of thermodynamics
Q = ∆U + W
= 0 + W
Q = W

v) Work done (W)
Work done = hatched area in PV diagram = height x breadth
W = 𝐏𝟏𝐕𝟏𝐥𝐧
𝐕𝟐
𝐕𝟏
(or) W = mR 𝐓𝟏 ln
𝐕𝟐
𝐕𝟏
∆S = 𝐓𝟏
𝐓𝟐 𝐝𝐐
𝐓
T is constant
=
𝟏
𝐓
𝒅𝑸
𝒅𝒙 = x
∆S =
𝐐
𝐓
=
mR 𝐓𝟏ln 𝐕𝟐
𝐕𝟏
𝐓𝟏
∆S = mR ln
𝐕𝟐
𝐕𝟏

• It is expanded rapidly.
• To prevent heat transfer, system is fully
covered by insulation.
• No heat transfer takes place during the process.
• So the entropy of the gas remains constant.

i) Pressure, Volume, Temperature
Relation:
i) Pressure Volume Relations
𝐏𝟏
𝐏𝟐
= [
𝐕𝟐
𝐕𝟏
]𝛄
ii. Temperature Volume Relations
𝐓𝟏
𝐓𝟐
= [
𝐕𝟐
𝐕𝟏
]𝛄−𝟏
ii. Temperature Pressure Relations
𝐓𝟏
𝐓𝟐
= [
𝐏𝟏
𝐏𝟐
]
𝜸−𝟏
𝜸

Q = 0
iii) Work done (W)
W =
𝑷𝟏𝑽𝟏−𝑷𝟐𝑽𝟐
𝜸−𝟏
∆U = m 𝐂𝐯 (T2 – T1)
v) By first law of thermodynamics
Q = ∆U + W
0 = ∆U + W
∆U = - W
∆S = 𝐓𝟏
𝐓𝟐 𝐝𝐐
𝐓
Q is 0
∆S = 𝟎

i) Pressure, Volume, Temperature
Relation:
i) Pressure Volume Relations
𝐏𝟏
𝐏𝟐
= [
𝐕𝟐
𝐕𝟏
]𝒏
ii. Temperature Volume Relations
𝐓𝟏
𝐓𝟐
= [
𝐕𝟐
𝐕𝟏
]𝐧−𝟏
ii. Temperature Pressure Relations
𝐓𝟏
𝐓𝟐
= [
𝐏𝟏
𝐏𝟐
]
𝒏−𝟏
𝒏

Q =
𝛄−𝐧
𝛄−𝟏
x W
iii) Work done (W)
W =
𝐏𝟏𝐕𝟏−𝐏𝟐𝐕𝟐
𝐧−𝟏
∆U = m 𝐂𝐯 (T2 – T1)
v) Change in Entropy (∆S)
∆S =
𝛄−𝐧
𝛄−𝟏
mR ln
𝐕𝟐
𝐕𝟏

Isothermal Process Isentropic Process
Temperature is Constant Entropy is Constant
Heat Transfer takes place No Heat Transfer takes place
Slow Process Fast Process
Change in Internal Energy is
Zero
Change in Internal Energy is
not Zero
Change in enthalpy is Zero Change in enthalpy is not Zero
Ex: Stirling Cycle Ex: Otto cycle

Problems on
Thermodynamic
Processes

Problems on Constant
Volume Processes

1. Determine the final temperature, external work done, change in
internal energy, change in enthalpy, change in entropy in the case of
2kg of gas at 50°C being heated at constant volume until the
pressure is doubled.
Given Data:
• Type of Process:
• Mass, m = 2 kg
• Initial Temperature, T1 = 50°C = 50 + 273 = 323 K
• Pressure is Doubled
To Find:
• Final Temperature, T2
• Work Done, W
• Change in Internal Energy, ∆U
• Change in Enthalpy, ∆h
• Change in Entropy, ∆S
40
Constant Volume Process
𝐏𝟐= 2 𝑷𝟏
6 April 2022 U.Aravind, Lect/Mech, LAPC

Solution:-
I. To Find Final Temperature, 𝐓𝟐
From General gas equation :
𝐏𝟏𝐕𝟏
𝐓𝟏
=
𝐏𝟐𝐕𝟐
𝐓𝟐
𝐏𝟏
𝐓𝟏
=
𝐏𝟐
𝐓𝟐
∴ 𝐓𝟐 =
𝐏𝟐𝐓𝟏
𝐏𝟏
𝑇2 =
2P1x 323
P1
= 646
41
𝐓𝟐 = 646 K
𝐅𝐨𝐫 𝐂𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐕𝐨𝐥𝐮𝐦𝐞 𝐏𝐫𝐨𝐜𝐞𝐬𝐬
𝐕𝟏 = 𝐕𝟐

II. To Find Work done, W
III. To Find Change in Internal Energy,
∆U
We Know ∆U = m 𝐂𝐕 (T2 – T1)
= 2 x 0.718 (646 - 323)
= 463.82
42
W = 0
𝐀𝐬𝐬𝐮𝐦𝐞, CV = 0.718
∆U = 463.82 kJ

IV. To Find Change in Enthalpy, ∆h
We Know ∆h = m 𝐂𝐏 (T2 – T1)
= 2 x 1.005 (646 - 323)
= 649.23
V. To Find Change in Entropy, ∆S
We Know ∆S = m𝐂𝐯 ln
𝐓𝟐
𝐓𝟏
= 2 x 0.718 ln
𝟔𝟒𝟔
𝟑𝟐𝟑
= 0.9953
43
𝐀𝐬𝐬𝐮𝐦𝐞, CP = 1.005
∆𝐡 = 649.23 kJ
∆𝐒 = 0.9953 kJ/K

Results:-
I. Final Temperature, 𝐓𝟐 = 646 K
II. Work done, W = 0
III. Change in Internal Energy, ∆U = 463.82 kJ
IV. Change in Enthalpy, ∆h = 649.23 kJ
V. Change in Entropy, ∆S = 0.9953 kJ/K
44

2. A Mass of 2.25 kg of nitrogen occupying 1.5𝐦𝟑
is heated from
25°C to 200°C at constant volume. Molecular weight of nitrogen is
28 and universal gas constant is 8.314 KJ/kg.mole.K. Calculate the
initial and final pressure of the gas and Change in internal energy
and change in entropy.
Given Data:
• Mass, m = 2.25 kg
• Volume, V1 = V2 = 1.5m3
• Initial Temperature, T1 = 25°C= 25 + 273 = 298 K
• Final Temperature, T2 = 200°C= 200 + 273 = 473 K
• Molecular Weight, M = 28
• Universal Gas Constant, RU = 8.314 KJ/kg.K
To Find:
i) Initial Pressure, P1 ii) Final Pressure, P2
iii) `Change in Internal Energy, ∆U iv)Change in Entropy, ∆S 45
Constant Volume Process

Solution:-
I. To Find Initial Pressure, 𝐏𝟏
From Characteristic gas equation :𝐏𝟏𝐕𝟏 = 𝐦𝐑𝐓𝟏
We have to find R
We know R =
RU
M
=
8.314
28
R = 0.2969
∴ 𝐏𝟏 =
𝐦𝐑𝐓𝟏
𝐕𝟏
=
2.25 x 0.2969 x 298
1.5
= 132.714
46
𝐏𝟏 = 132.714 kN/ 𝐦𝟐

II. To Find Final Pressure, 𝐏𝟐
From Characteristic gas equation :𝐏𝟐𝐕𝟐 = 𝐦𝐑𝐓𝟐
∴ 𝐏𝟐 =
𝐦𝐑𝐓𝟐
𝐕𝟐
=
2.25 x 0.2969 x 473
1.5
= 210.65
III. To Find Change in Internal Energy, ∆U
= 2.25 x 0.718 (473-298)
= 282.71
47
𝐏𝟐= 210.65 kN/ 𝐦𝟐
∆U = 282.71 kJ

IV. To Find Change in Entropy, ∆S
We Know ∆S = m𝐂𝐯 ln
𝐓𝟐
𝐓𝟏
= 2.25 x 0.718 ln
473
298
= 0.7463
Results:-
I. Initial Pressure, 𝐏𝟏 = 132.714 kN/ 𝐦𝟐
II. Final Pressure, 𝐏𝟐 = 210.65 kN/ 𝐦𝟐
III. Change in Internal Energy, ∆U =282.71 kJ
IV.Change in Entropy, ∆S =0.7463 kJ/K
48
∆𝐒 = 0.7163 kJ/K

Problems on Constant
Pressure Process

3. A gas whose pressure, volume and temperature are 5 bar,
0.23𝐦𝟑and 185°C respectively has changed its state at constant
pressure until its temperature becomes 70°C. Determine (i) Mass (ii)
Work done (iii) Change in Internal Energy (iv) Heat Transferred
during the process. Take R = 0.29 kJ/kg.K and CP = 1.005 kJ/kg. K
Given Data:
• Pressure, P1 = P2 = 5 bar = 5 x 100 = 500 kN/ m2
• Initial Volume, V1 = 0.23 m3
• Final Temperature, T2 = 70°C= 70 + 273 = 343 K
• Gas Constant, R = 0.29 KJ/kg.K
• CP = 1.005 kJ/kg. K
To Find:
i) Mass, M ii) Work done, W
iii) Change in Internal Energy, ∆U iv) Heat Transferred, Q 50
Constant Pressure Process

Solution:-
I. To Find Mass, m
∴ m =
𝐏𝟏𝐕𝟏
𝐑𝐓𝟏
=
500 x 0.23
0.29 x 458
= 0.8658
II. To Find Change in Internal Energy, ∆U
= 0.8658 x 0.718 (343-458)
= - 71.19 kJ
51
𝐦 = 𝟎. 𝟖𝟔𝟓𝟖 𝐤𝐠
∆U = −71.19 kJ

III. To Find Heat Transferred, Q
We Know Q= m 𝐂𝐏 (T2 – T1)
= 0.8658 x 1.005 (343-458)
= - 100.064 kJ
IV. To Find Work done, W
By First law of Thermodynamics
Q = ∆U + W
∴ 𝐖 = Q - ∆U
= - 100.064 - (-71.19)
= -28.874
52
𝐐 = −𝟏𝟎𝟎. 𝟎𝟔𝟒 𝐤𝐉
𝐖 = −28.874 kJ

4. One kg of air at a pressure of 5 bar and volume 0.3𝐦𝟑
and
temperature 627°C is supplied with heat at constant pressure till the
volume is doubled. Calculate (i) Work done (ii) Change in Internal
Energy (iii) Heat Transferred (iv) Change in Entropy (v) Change in
Enthalpy. Take R = 0.287 kJ/kg.K and CP = 1.005 kJ/kg. K , Cv =
0.718 kJ/kg. K
Given Data:
• Mass of air, m = 1kg
• Pressure, P1 = P2 = 9 bar = 9 x 100 = 900 kN/ m2
• Initial Volume, V1 = 0.3 m3
• Volume is Doubled
• Gas Constant, R = 0.29 KJ/kg.K
• CP = 1.005 kJ/kg. K
• Cv = 0.718 kJ/kg. K
53
Constant Pressure Process
𝐕𝟐= 2 𝐕𝟏 = 2 x 0.3 = 0.6 m3

To Find:
• Work Done, W
• Heat Transferred, Q
Solution:-
I. To Find Final Temperature, 𝐓𝟐
From General gas equation :
𝐏𝟏𝐕𝟏
𝐓𝟏
=
𝐏𝟐𝐕𝟐
𝐓𝟐
𝐕𝟏
𝐓𝟏
=
𝐕𝟐
𝐓𝟐
∴ 𝐓𝟐 =
𝐕𝟐𝐓𝟏
𝐕𝟏
T2 =
0.6 x 900
0.3
= 1800 54
𝐅𝐨𝐫 𝐂𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐏𝐫𝐞𝐬𝐬𝐮𝐫𝐞 𝐏𝐫𝐨𝐜𝐞𝐬𝐬
𝑷𝟏 = 𝑷𝟐
𝐓𝟐 = 𝟏𝟖𝟎𝟎 𝐊

II. To Find Work done
We know :𝐖 = 𝐏𝟏(𝐕𝟐 - 𝐕𝟏)
= 900 (0.6-0.3)
= 270
III. To Find Change in Internal Energy, ∆U
= 1 x 0.718 (1800-900)
= 646.2
55
∆U = 646.2 kJ
W = 𝟐𝟕𝟎 kJ

IV. To Find Change in Entropy, ∆S
We Know ∆S = m𝐂𝐏 ln
𝐓𝟐
𝐓𝟏
= 1 x 1.005 ln
1800
900
= 0.6969
V. To Find Change in Enthalpy , ∆h
= 1 x 1.005 1800 − 900
= 904.5
56
∆𝐒 = 0.6969 kJ/K
∆𝐡 = 904.5 kJ

VI. To Find Heat Transferred, Q
We Know Q= m 𝐂𝐏 (T2 – T1)
= 1 x 1.005 1800 − 900
= 904.5
57
𝐐 = 904.5 kJ

Problems on Constant Temperature
Process (or) Isothermal Process

5. 0.2 kg of air at a pressure of 1.1 bar and 15°C is
compressed to isothermally to a pressure of 5.5 bar.
Calculate (i) Initial & Final Volume (ii) Work done (iii)
Change in Internal Energy (iv) Heat Rejected (v)
Change in Entropy (vi) Change in Enthalpy. Take R =
0.292 kJ/kg.K
Given Data:
•Type of Process:
•Mass of air, m = 0.2 kg
•Initial Pressure, P1 = 1.1 bar = 1.1 x 100 = 110 kN/ m2
•Final Pressure, P2 = 5.5 bar = 5.5 x 100 = 550 kN/ m2
•Temperature, T1 = T2 = 15°C= 15 + 273 = 288 K
•Gas Constant, R = 0.29 KJ/kg.K
59
Constant Thermal Process

To Find:
• Initial Volume, V1
• Final Volume, V2
• Work Done, W
• Heat Rejected, Q
Solution:-
I. To Find Initial Volume, 𝐕𝟏
𝐕𝟏 =
𝐦𝐑𝐓𝟏
𝐏𝟏
𝐕𝟏 =
0.2 x 0.292 x 288
110
= 0.1529
60
𝐅𝐨𝐫 𝐂𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐓𝐞𝐦𝐩𝐞𝐫𝐚𝐭𝐮𝐫𝐞 𝐏𝐫𝐨𝐜𝐞𝐬𝐬
𝐓𝟏 = 𝐓𝟐
𝐕𝟏 = 𝟎. 𝟏𝟓𝟐𝟗 𝐦𝟑

II. To Find Final Volume, 𝐕𝟐
From Characteristic gas equation :𝐏𝟐𝐕𝟐 = 𝐦𝐑𝐓𝟐
𝐕𝟐 =
𝐦𝐑𝐓𝟐
𝐏𝟐
𝐕𝟐 =
0.2 x 0.292 x 288
550
= 0.0305
III. To Find Work done, W
We know :W = 𝐏𝟏𝐕𝟏𝐥𝐧
𝐕𝟐
𝐕𝟏
= 110 x 0.1529 ln
0.0305
0.1529
= - 27.113
61
𝐕𝟐 = 𝟎. 𝟎𝟑𝟎𝟓 𝐦𝟑
𝐖 = −27.113 kJ

IV. To Find Change in Internal Energy, ∆U
= 0.2 x 0.718 (288-288)
= 0
V. To Find Change in Enthalpy , ∆h
= 0.2 x 1.005 288 − 288
= 0
62
∆U = 0
∆𝐡 = 𝟎

V. To Find Change in Entropy, ∆S
We Know ∆S = m𝐑ln
𝐕𝟐
𝐕𝟏
= 0.2 x 0.292 ln
0.0305
0.1529
= -0.094
VII. To Find Heat Transferred, Q
We know :Q = 𝐏𝟏𝐕𝟏𝐥𝐧
𝐕𝟐
𝐕𝟏
= 110 x 0.1529 ln
0.0305
0.1529
= - 27.113
63
∆𝐒 = −0.094 kJ/K
𝐐 = −27.113 kJ

Problems on Adiabatic Process
(or) Isentropic Process

6. Air at 1 bar and 40°C is compressed to 1/10th original
volume isentropically. Calculate Final pressure and
Temperature & Work done on 1𝐦𝟑of air. Assume R =
0.287 kJ/kg.K and γ = 1.41
Given Data:
•Type of Process:
•Mass of air, m = 1kg
•Initial Pressure, P1 = 1 bar = 1 x 100 = 100 kN/ m2
•Initial Temperature, T1 = 40°C= 40 + 273 = 313 K
•Initial Volume, V1 = 1 m3
•Final volume V2 =
•Gas Constant, R = 0.29 KJ/kg.K
•γ = 1.41 65
Isentropic Process
1/10th 𝐭𝐡𝐞 𝐨𝐫𝐢𝐠𝐢𝐧𝐚𝐥 𝐯𝐨𝐥𝐮𝐦𝐞
=
1
10
x V1
=
1
10
x 1 = 0.1m3

To Find:
• Final Pressure, P2
• Work Done, W
Solution:-
I. To Find Final Pressure, P2
Pressure Volume Relationship:
Writing Inversely,
∴ P2 = P1 x
V1
V2
γ
= 100 x
1
0.1
1.41
= 2570.395
66
𝐏𝟏
𝐏𝟐
=
𝐕𝟐
𝐕𝟏
𝛄
𝐏𝟐
𝐏𝟏
=
𝐕𝟏
𝐕𝟐
𝛄
𝐏𝟐 = 2570.395 kN/ 𝐦𝟐

II. To Find Final Temperature, T2
Pressure Volume Relationship:
Writing Inversely,
∴ T2 = T1 x
V1
V2
γ−1
= 313 x
1
0.1
1.41−1
= 804.53
67
𝐓𝟏
𝐓𝟐
=
𝐕𝟐
𝐕𝟏
𝛄−𝟏
𝐓𝟐
𝐓𝟏
=
𝐕𝟏
𝐕𝟐
𝛄−𝟏
𝐓𝟐 = 804.53 K

III. To Find Work done, W
We know :
=
(100 x 1) – (2570.395 x 0.1)
1.41 − 1
= -383.02
68
W =
𝐏𝟏𝐕𝟏−𝐏𝟐𝐕𝟐
𝛄−𝟏
W = -383.02 kJ

7. 0.5 kg of a certain gas occupies 0.3 𝐦𝟑
at 20°C
and 140 kN/𝐦𝟐
and after adiabatic compression to
0.15𝐦𝟑
, the pressure is 370 kN/𝐦𝟐
. Find the value
of gas constant and two specific heats. Take γ = 1.4
Given Data:
•Type of Process:
•Mass of air, m = 0.5 kg
•Initial Volume, V1 = 0.3 m3
•Initial Temperature, T1 = 20°C= 20 + 273 = 293 K
•Initial Pressure, P1 = 140 kN/m2
•Final volume V2 = 0.15 m3
•Final Pressure, P2 = 370 kN/m2
• γ = 1.4 69
Isentropic Process
U.Aravind, Lect/Mech, LAPC

To Find:
•Gas Constant, R
•Specific Heat, CP
•Specific Heat, Cv
Solution:-
I. To Find Gas Constant, R
∴ R =
P1V1
mT1
=
140 x 0.3
0.5 x 293
= 0.2866
70
R = 0.2866 kJ/kg.K

II. To Find Specific Heat, 𝐂𝐯
We know :𝐏𝟏𝐕𝟏 = 𝐦𝐑𝐓𝟏
∴ R =
P1V1
mT1
=
140 x 0.3
0.5 x 293
= 0.2866
71
R = 0.2866 kJ/kg.K

Thermal Engineering - I - Unit 1 Thermodynamics Process of Perfect Gases - N Scheme III Sem Diploma Mech

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