Eme ch 3_properties_of_gases_ppt

Elements of Mechanical Engineering

Chapter 3 Properties of Gases

Prepared by :- Mr. Mitesh D. Gohil

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Outline

3.1 Gas Laws like Boyle’s Law, Charle’s law, Gay lussac law

3.2 Combined gas law or Ideal gas equation

3.3 Basic gas processes

3.4 General Procedure for Applying the first law of thermodynamics to

different processes

3.5 Constant Volume Process / Isochoric process in Close System

3.6 Constant Pressure Process / Isobaric Process in Close System

3.7 Constant Temperature Process / Isothermal Process / Hyperbolic Process

3.8 Reversible adiabatic Process

3.9 Polytropic Process

3.10 The combination of Polytropic law (𝐏𝐕𝛄 = 𝐂𝐨𝐧𝐬𝐭. ) and Equation of State

(PV = mRT)

Ideal Gas:-

Ideal gas is “which would behave in an ideal manner at all pressure

and temperature such a gas will follow all gas laws and

characteristics gas equation.

Real Gas:-

The behavior of all real gas at high pressure and low temperature is

different and in that condition real gas will not follow gas laws.

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3.1 Gas Laws

4

P ∝1

V(T= Constant)

PV = Constant

P1V1 = P2V2 = Constant …(1)

Boyle’s law

The absolute pressure of a given mass of an ideal gas is inversely

proportional to its volume, if the temperature of the gas is kept constant.

3.1 Gas Laws… continue

Q. State the following : Boyles Law, Charle’s Law GTU : April 2010

Fig. P-V diagram for constant

temperature process from state 1 to

2 in close system

5

Charles’s Law

Volume of a given mass of an ideal gas is directly proportional to its absolute

temperature, if the pressure of gas is kept constant.

V ∝ T (P= Constant)

V

T= Constant

V1T1=V2T2= Constant …(2)

Gay lussac law

The absolute pressure of a given mass of an ideal gas is proportional to

temperature, if the volume of gas is kept constant.

P ∝ T (V= Constant)

P

T= Constant

P1T1=P2T2= Constant …(3)


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Avogadro’s law Avogadro’s law states that equal volume of different ideal gases at the

same pressure and temperature contain the same number of molecules.

OR

Avogadro’s law states that the volume of a g mol of all gases at the

pressure of 1 atm. and temperature of 0° C is the same, and equal to 22.4

liters.

1 kg mole contains 6.023 × 1023 molecules.

A mole (n) of a substance has numerically equal to the molecular weight

(M) of the substance.

Example:- 32 gm 𝑂2 = 1 g mole

28 kg 𝑁2= 1 kg mole

Mole of gas, n

n =m

M,kg moles …(4)

Where , m = mass of substance, kg

M= Molecular weight, kg/kg mole


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3.2 Combine gas law

Combine gas law / Characteristic gas equation / Equation of state / Ideal gas

equation

For an ideal gas, the equation of state represents the relationship between pressure

(p), volume (V) and Temperature (T).

𝑓 𝑃, 𝑉, 𝑇 = 0

It is possible to change the state 1 to

state 2 by infinite number of paths.

However we will follow path 1-3-2.

Because in that path, from 1-3 there is

constant volume process, in that we can

apply gay lussac law and from 3-2 there

is constant pressure process, in that we

can apply charl’s law.

Q. State the following : Characteristic gas equation. GTU : April 2010

Q. Derive characteristic equation of a perfect gas. GTU : Dec 2008

Fig. P-V diagram for any process from state

1 to 2 in close system

8

P1V1T1

=P2V2T2

= constant(∵ V3 = V1 & P3 = P2) …(7)

Put the value of T3 from Eq. (5) to (6)

V3P3T1P1

=V2T2

P1V3T1

=P3V2T2

For Process 1-3

P1T1=P3T3

(Gay lussac law) …(5)

For Process 3-2

V3T3=V2T2

(Charles’s law) …(6)

Fig. P-V diagram for any process

from state 1 to 2 in close system

3.2 Combine gas law… continue

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So, for any fix mass of gas (close system), the change of state is connected by equation

𝑃𝑉

𝑇= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑃𝑉

𝑇= 𝑚𝑅

𝑃𝑉 = 𝑚𝑅𝑇 …(8)

𝑃𝑣 = 𝑅𝑇 ∵ 𝑣 =𝑉

𝑚…(9)

Where , R= Characteristic gas constant

𝑣 = 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑉𝑜𝑙𝑢𝑚𝑒

Equation (8) is known as Combine gas law / Characteristic gas equation /

Equation of state / Ideal gas equation for an ideal gas.

Q. Derive Expression pV/T = constant with help of Boyle’s law and Charles’s law. GTU : May 2013


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R depends on the kind of gas.

Example, For air, R= 287 Nm

kg K

Significance of R

R represents the magnitude of work that can be done by 1 kg of gas when subjected to

1 degree change in temperature.

Putting the value of m in terms of mole (n) in equation (8)

Where, R = Universal gas constant= 8.31 kJ/ (kg mol) (K)

R does not depends on the kind of gas and it is same for all gases.

Real gases are not exactly obey this equation of state for all p, V, T value.

Real gases at very low pressure and very high temperature will not obey this equation

of state.

PV = nMRT ∵ n =m

M…(10)

PV = n RT (Putting, MR = R ) …(11)


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Application of first law of thermodynamics as applied to closed system

1. Constant Volume Process / Isochoric process,

V = Constant

2. Constant Pressure Process / Isobaric Process,

P = Constant

3. Constant Temperature Process / Isothermal Process / Hyperbolic Process,

T = constant

4. Reversible adiabatic Process,

Q = 0, or PVγ = Constant

Where γ = Adiabatic index

5. Polytropic Process,

PVn = Constant

Where n = Polytropic index

Q. What are basic gas processes ? GTU : May 20123.3 Basic gas processes

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Fig. P-V Diagram for different basic gas processes

3.3 Basic gas processes … continue

Q. How the basic processes are shown graphically on P-v diagram ? GTU : May 2012

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3.4

1) Work done, W= P dV

2) Heat Supply, Q

δQ = m Cp or v dT

Q = m Cp or v ∆T

3) Change in Internal Energy, ∆U

dU = m CvdT

∆U = mCv∆T

4) Application of 1st law of thermodynamic,

δQ − δW = dU

Q −W = ∆U

Above equation could be derived from constant volume process and true for all processes.

Where, 𝐶𝑝 = Specific heat at constant pressure

𝐶𝑣 = Specific heat at constant volume

…(13)

…(14)

…(15)

…(16)

General Procedure for applying the 1st law of thermodynamics to closed

system

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5) Change in Enthalpy, ∆H

h = u + pv

dh = du + p dv + v dp (differentiated above equation with respect to h)

6) Equation of state,

PV = mRT

7) Relation between Cp and Cv,

Cp − Cv = R

Equation Cp − Cv = R, could be derived from constant pressure process and true

for all processes.

Cp

Cv= γ

Where, H = Enthalpy (J)

h = Specific Enthalpy (J/kg)

u = Internal energy (J)

Where, 𝐶𝑝 = Specific heat at constant pressure

𝐶𝑣 = Specific heat at constant volume

𝛾 𝑖𝑠 𝑎𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐 𝑖𝑛𝑑𝑒𝑥

…(17)

…(18)

…(19) …(20)

3.4General Procedure for applying the 1st law of thermodynamics to closed

system … continue

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3.5 Constant Volume Process / Isochoric process in Close System

State 1 :- P1, V1, T1

State 2 :- P2, V2, T2

V1 = V2

V = Constant

dV = 0

1) Work done, W= P dV = 0

2) Heat Supply, Q

δQ = m Cv dT

Q = m Cv ∆T

Q = m Cv (T2 − T1)

…(21)

…(22)Fig. P-V diagram for constant

volume process from state 1 to 2 in

close system

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δQ − δW = dU

Q −W = ∆U

m Cv T2 − T1 − 0 = ∆U

∆U = m Cv T2 − T1

Above equation (23) is derived from constant volume process and true

for all processes.

For infinitesimal process, change in internal energy, 𝑑𝑢 = 𝑚 𝐶𝑣 𝑑𝑇

For unit mass, Cv =du

dT v

Another word,

du ∝ dT

So, Internal energy of substance is function of temperature only.

…(23)

3.5 Constant Volume Process / Isochoric process in Close System…cont.

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3.6 Constant Pressure Process / Isobaric Process,

P = Constant


pressure process from state 1 to 2

in close system

State 1 :- 𝑃1, 𝑉1, 𝑇1

State 2 :- 𝑃2, 𝑉2, 𝑇2

𝑃1 = 𝑃2 = 𝑃

1) Work done,

W =

1

2

𝑃 𝑑𝑉

= 𝑃

1

2

𝑑𝑉 (∵P= constant)

= 𝑃 𝑉 12

= 𝑃 𝑉2 − 𝑉1

= 𝑃2𝑉2 − 𝑃1𝑉1 (∵ 𝑃1= 𝑃2 = 𝑃)

= 𝑚𝑅𝑇2 −𝑚𝑅𝑇1 (∵ PV = mRT)

= 𝑚𝑅 𝑇2 − 𝑇1 …(24)

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𝑊 = 𝑚𝑅 𝑇2 − 𝑇1

3.6 Constant Pressure Process / Isobaric Process, …Continue

2) Heat Supply, Q

𝛿𝑄 = 𝑚 𝐶𝑝 𝑑𝑇

𝑄 = 𝑚 𝐶𝑝 (𝑇2 − 𝑇1)

3) Change in Internal Energy, ∆𝑈

𝑑𝑈 = 𝑚 𝐶𝑣𝑑𝑇

∆U = mCv(T2 − T1)


𝛿𝑄 − 𝛿𝑊 = 𝑑𝑈

𝑄 −𝑊 = ∆𝑈

𝑚 𝐶𝑝 𝑇2 − 𝑇1 − 𝑚𝑅 𝑇2 − 𝑇1 = mCv(T2 − T1)

𝐶𝑝 − Cv = 𝑅

Above equation (27) is called Meyer’s Equation

..(25)

(From eq.(14))

..(26)

..(27)

Q. With usual notation prove that Cp – Cv = R GTU : June 2009

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For another relationship

𝑄 −𝑊 = ∆𝑈

𝑄 = ∆𝑈 +𝑊

= 𝑈2 − 𝑈1 + 𝑃2𝑉2 − 𝑃1𝑉1= 𝑈2 + 𝑃2𝑉2) − (𝑈1+𝑃1𝑉1= 𝐻2 −𝐻1

Heat transfer in a constant pressure process is equal to change in enthalpy of gas.

Another word,

𝑑ℎ ∝ 𝑑𝑇

So, enthalpy of substance is function of temperature only.

3.6 Constant Pressure Process / Isobaric Process, …Continue

m Cp dT = dH (from equation (25))

dh = Cp dT

Cp =dh

dTp

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3.7 Constant Temperature Process / Isothermal Process

T = constant


temperature process from state 1

to 2 in close system

State 1 :- 𝑃1, 𝑉1, 𝑇1

State 2 :- 𝑃2, 𝑉2, 𝑇2

𝑇1 = 𝑇2 = 𝑇

PV=mRT (from the equation of state)

PV=constant

(∵ In close system mR = const. & T = const.)

PV=C

∴ 𝑃1 𝑉1 = 𝑃2𝑉2 = 𝐶

So, this process follow the Boyle’s law

Q. Explain Isothermal Process. For Isothermal process, Find expression of work done,

Change in internal Energy, Change in Enthalpy and Heat transfer.

GTU : Dec 2012

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3.7 Constant Temperature Process / Isothermal Process….Continue

1) Work done, W

W =

1

2

𝑃 𝑑𝑉 (from eq. (13)

=

1

2𝐶

𝑉𝑑𝑉 (∵PV= C)

= 𝐶 ln𝑉 12

= 𝐶 ln𝑉2 − ln𝑉1

= 𝐶 ln𝑉2𝑉1

= 𝑃1𝑉1 ln𝑉2𝑉1

(∵ 𝑃1𝑉1 = 𝑃2𝑉2 = 𝐶)

= 𝑚𝑅𝑇1 ln𝑉2𝑉1

(∵ PV = mRT)

W = 𝑚𝑅𝑇1 ln𝑃1𝑃2

(∵ 𝑃1𝑉1 = 𝑃2𝑉2) …(28)

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2) Change in Internal Energy, ∆𝑼

dU = m CvdT

∆U = mCv∆T (From equation (15))

∆U = 0 (∆T = 0, Isothermal Process) …(29)



𝑄 −𝑊 = ∆𝑈 (From equation (16))

𝑄 = ∆𝑈 +𝑊

𝑄 = 𝑚𝑅𝑇1 ln𝑃1𝑃2

(From (28)&(29) …(30)

4) Change in Enthalpy, ∆𝑯

h = u + pv (From equation (17))

H=U+PV

𝐻2 − 𝐻1 = (𝑈2 − 𝑈1) + (𝑃2𝑉2 − 𝑃1𝑉1)

𝐻2 − 𝐻1 = 0 (∵∆𝑈 = 0 & 𝑃1𝑉1 = 𝑃2𝑉2) …(31)

3.7 Constant Temperature Process / Isothermal Process….Continue

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3.8 Reversible adiabatic Process,

𝑄 = 0, & 𝑃𝑉𝛾 = Const.

Q. Prove that 𝑷𝑽𝜸 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕, for an adiabatic process.

( GTU : Dec 2010, June 2011, Dec 2011)

In Adiabatic process we know, 𝑄 = 0 𝑜𝑟 𝛿𝑄 = 0

Now From State of equation,

𝑃𝑉 = 𝑚𝑅𝑇

𝑃𝑑𝑉

𝑑𝑇+ 𝑉

𝑑𝑃

𝑑𝑇= 𝑚𝑅

(differentiated above equation

with respect to T)

𝑃𝑑𝑉 + 𝑉𝑑𝑃 = 𝑚𝑅𝑑𝑇 …(32)

From the 1st law of thermodynamic,


𝑑𝑈 + 𝛿𝑊 = 0 (∵𝛿𝑄 = 0)

𝑚𝐶𝑣𝑑𝑇 + 𝑃𝑑𝑉 = 0 (from eq.(13) &(15)) …(33)

𝑑𝑇 =−𝑃𝑑𝑉

𝑚𝐶𝑣

Where γ = Adiabatic index

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3.8 Reversible adiabatic Process, ….Continue

𝑃𝑑𝑉 + 𝑉𝑑𝑃 = 𝑚𝑅−𝑃𝑑𝑉

𝑚𝐶𝑣

𝑃𝑑𝑉 + 𝑉𝑑𝑃 =𝑅

𝐶𝑣−𝑃𝑑𝑉

𝑃𝑑𝑉 + 𝑉𝑑𝑃 =𝐶𝑝 − 𝐶𝑣

𝐶𝑣−𝑃𝑑𝑉 (∵𝑅 = 𝐶𝑝 − 𝐶𝑣)

𝑃𝑑𝑉 + 𝑉𝑑𝑃 = 𝛾 − 1 −𝑃𝑑𝑉 (∵𝐶𝑝

𝐶𝑣= 𝛾)

𝛾 𝑃𝑑𝑉 + 𝑉𝑑𝑃 = 0

𝛾𝑑𝑉

𝑉+𝑑𝑃

𝑃= 0

𝛾 ln𝑉 + ln𝑃 = ln𝐶(Integrating the above

equation)

ln 𝑉𝛾 + ln𝑃 = ln𝐶

ln 𝑃𝑉𝛾 = ln𝐶

𝑃𝑉𝛾 = 𝐶 …(34)

Substituting the value of dT in eq. (32) from eq. (33)

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1) Work done, W

𝑊 =

1

2

𝑃 𝑑𝑉 (Form eq.(13))

=

1

2

𝐶𝑉−𝛾 𝑑𝑉 (∵ 𝑃𝑉𝛾 = 𝐶)

= 𝐶𝑉1−𝛾

1 − 𝛾1

2

=𝐶

1 − 𝛾𝑉21−𝛾 − 𝑉1

1−𝛾

=1

1 − 𝛾𝐶𝑉2

1−𝛾 − 𝐶𝑉11−𝛾

=1

1 − 𝛾𝑃2𝑉2 − 𝑃1𝑉1 (∵ 𝑃2𝑉2

𝛾 = 𝑃1𝑉1𝛾 = 𝐶)

=1

1 − 𝛾𝑚𝑅𝑇2 −𝑚𝑅𝑇1 ∵ 𝑃𝑉 = 𝑚𝑅𝑇

=𝑚𝑅 𝑇2 − 𝑇11 − 𝛾

…(35)W


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𝑄 −𝑊 = ∆𝑈

𝑊 = −∆𝑈 (∵Q = 0)

𝑊 = mCv(𝑇1 − 𝑇2) (∵ ∆U = mCv 𝑇2 − 𝑇1 ) …(36)


Q. Define adiabatic process. Derive the relation between P, V, T for this process. Also

derive the expression for work done and change in internal energy for this process.

GTU : Jan 2011

27

3.9 Polytropic Process,

PVn = Constant

Where n = Polytropic index

1) Work done, W

W =

1

2

P dV (Form eq.(13))

= 12CV−n dV (∵ PVn = C)

= CV1−n

1 − n1

2

=C

1 − nV21−n − V1

1−n

=1

1 − nCV2

1−n − CV11−n

=1

1 − nP2V2 − P1V1 (∵ P2V2

n = P1V1n = C)

=1

1 − nmRT2 −mRT1 ∵ PV = mRT

W =mR T2 − T11 − n

…(37)

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2) Change in Internal Energy, ∆𝐔

dU = m CvdT (from eq.(15))

∆U = mCv T2 − T1 …(38)


δQ − δW = dUQ = W+ ∆U

=mR T2 − T11 − n

+mCv T2 − T1 (From eq. 37 &(38))

=m Cp − Cv T2 − T1

1 − n+mCv T2 − T1

(∵R = Cp − Cv)

= m Cv

CpCv− 1

1 − n+ 1 T2 − T1

= m Cv𝛾 − 1

1 − n+ 1 T2 − T1

(∵Cp

Cv= γ)

Q = m Cv𝛾 − n

1 − nT2 − T1

…(39)

3.9 Polytropic Process,….Continue

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𝑄 = 𝑚 𝐶𝑛 𝑇2 − 𝑇1 (Putting, 𝐶𝑣𝛾−𝑛

1−𝑛= 𝐶𝑛)

Where, 𝐶𝑛=Polytropic specific heat

3.9 Polytropic Process,….Continue

From eq. (39)

30

3.10 Combination of Polytropic law (𝐏𝐕𝛄 = 𝐂𝐨𝐧𝐬𝐭. ) and Equation of State (PV = mRT)

Polytropic law (𝑃𝑉𝑛 = 𝐶𝑜𝑛𝑠𝑡. )

P1V1n = P2V2

n

∴P1P2

=V2V1

n

…(40)

Polytropic law (𝑃𝑉𝑛 = 𝐶𝑜𝑛𝑠𝑡. )

P1V1n = P2V2

n

P1V1 V1n−1

= P2V2 V2n−1

mRT1 V1n−1

= mRT2 V2n−1 ∵ PV = mRT

T1T2

=V2V1

n−1

…(41)

31

From the eq. (40)

From the eq. (41)

P1P2

1n

=V2V1

…(40)(a)

T1T2

1n−1

=V2V1

…(41)(b)

From the eq. (40)(a) & (41)(b)

T1T2

1n−1

=P1P2

1n

T1T2

=P1P2

n−1n …(42)

3.10 Combination of Polytropic law (𝐏𝐕𝛄 = 𝐂𝐨𝐧𝐬𝐭. ) and Equation of State (PV = mRT)…

Q. State the following : Boyles Law, Charle’s Law GTU : April 2010

Q. Derive characteristic equation of a perfect gas. GTU : Dec 2008

Q. State the following : Characteristic gas equation. GTU : April 2010

Q. Derive Expression pV/T = constant with help of Boyle’s law and Charles’s law. GTU : May 2013

Q. What are basic gas processes ? GTU : May 2012

Q. With usual notation prove that Cp – Cv = R GTU : June 2009

Q. Explain Isothermal Process. For Isothermal process, Find expression of work done,

Change in internal Energy, Change in Enthalpy and Heat transfer.

GTU : Dec 2012

Q. Prove that 𝑷𝑽𝜸 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕, for an adiabatic process. GTU : Dec 2010,

June 2011, Dec

2011

Q. Define adiabatic process. Derive the relation between P, V, T for this process. Also

derive the expression for work done and change in internal energy for this process.

GTU : Jan 2011

Q. How the basic processes are shown graphically on P-v diagram ? GTU : May 2012

Questions of GTU Exam

32

Thank You…

33