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L.VIJAYAKUMAR /A.P-MECH

MAHALAKSHMI

ENGINEERING COLLEGE

TIRUCHIRAPALLI- 621213.

QUESTION BANK

DEPARTMENT: MECH SEMESTER – III

SUBJECT CODE: EE2201 SUBJECT NAME: ENGINEERING THERMODYNAMICS

UNIT-IV- IDEAL AND REAL GASES, THERMODYNAMICS RELATIONS

(1)Write down the Dalton’s Law of Partial Pressure and Explain its importance

[AU DEC 05]

According to Dalton’s law of partial pressure, the pressure of gas mixture is equal to

the sum of pressure of its each components if each component is exerted alone of the

temp and volume of the mixture. This law is also called as Dalton’s law of Additive

pressure.

+ =

Where,

Pm= mixture pressure

P1 , P2 ,…………………Pi= Each component presser

Gas A

V, T

PA

Gas B

V, T

PB

Gas mix

A+B

V, T

PA+PB


If there are,

NA=moles of gas ‘A’

NB=moles of gas ‘B’

NC=moles of gas ‘C’

The gas equation is given by

Pm Vm = (NA+NB+NB) Tm

Where,

=8.3143 KJ/Kg mole k

Pm=PA(NA,Tm,Vm)+PB(NB,Tm,Vm)+Pc(NA,Tm,Vm)

Where,

PA(NA,Tm,VM)=> is the pressure of NA of component

Tm =Temperature

Vm= Volume

For ideal gas

For Real gas


Where,

Zm=compressibility Factor for the mixture

=>Zm can be Expressed in terms of compressibility factors the Individual

Gases Zi

Where,

Zi=is determined at Tm and Vm

(2)‘’Deive maxwell’s equation’’ [AU DEC 07]

The maxwell’s eqation realte entropy to the three directly heasurable properties P,V,andT

For pure simple compressible substances.

From first law of thermodynamics

Q=W+Δu

Rearranging the Parameters

Q= Δu+W

Tds = du +Pdv

Where,

ds= Qby second law of thermodyamics

W= Pdv by first law of thermodynamics

du = Tds – Pdv ---------------------------(1)

We known that,

H = u+Pv

dh = du+d(Pv)


dh = du+vdp+pdv -----------------------------(2)

Subsitituting the value of ‘’du’’ in eqution (2)

dh = Tds –Pdv + vdP + Pdn

dh = Tds + vdP ----------------------------(3)

By Helmotz’s function,

a = u – Ts

da = du – d(Ts)

da = du – Tds + sdT -----------------------------(4)

By differential rule d(uv) = udv + vdu sub the value of du in equation (4)

da = Tds – Pdv – Tds – sdT

da = -Pdv – sdT ------------------------------(5)

By Gibbs Function,

G = h – Ts

dg = dh – d(Ts)

dg = dh – Tds – sdT -------------------------------(6)

Sub th e value ‘’dh’’ in equation (6),

dg = Tas +vdp – Tds - sdT

dg = vdp – sdT -------------------------------(7)

By Inverse exact differential equation (1)

dU = TdS-PdV------------------(1)

[


Similarly equation(3) can be written as

dh =Tds+VdP------------------(3)

Similarly Equation (5) can be written as

da=-PdV-SdT

Similarly equation (7) can be written as

dg=VdP-SdT

These equation 8,9,10&11 are maxwell’s equations

(3)Derive the clausis-clapey Equation [AU DEC 07]

The clapeyron equation provides the relationship between the saturation pressure,

temperature, the enthalpy(lartent heat) of evaporation and the specific volume of the phases

involved.

This equation will be useful for finding the properties of pure substance exiting in two

phases in equilibrium.

It defines the slope of the curve separating the two phases in the p-t diagram.

Let, entropy(s) is a function of temp(T) and volume(V).

S=f (T,V)

When the phase is changing from saturated liquid to saturated vapor, temperature

remains constant.

So, ds equation reduces to


[Temp remains constant dT=0]

From Maxwell equation, we know that

Sub in equation

The term is the slope saturation curve

Integrating the above equation between saturated liquid (f) and saturated vapor (g)

----------------(3)

From second law of thermodynamics, we know that

For constant pressure process

dQ=dh


Sub in equation (3)

This equation is known as clapeyron equation.

(4) Derive the Vander waal’s equation [AU DEC 08,AU MAY 2011]

The scientist Vander waal’s considered these two correction to analyses the behaviors of

real gases during 1897

The equation of state for real gases is given by

The term a is to account for the molecular attraction

The term b(known as co volume) is to account for the volumes of the molecules

The terms is known as force of cohesion.


(5) Derive Tds equation taking temp, volume, and temp pressure a independent

properties [AU MAY 08]

The entropy (s) of a pure substance can be Expressed as a function of temp (T)

and pressure (P)

We known that

Sub in ds equation

multiplying by ‘T’ on both side of equation

This is known as the first form of entropy equation (or) The first TdS equation

By considering the entropy of a pure substance as a junction of temp and specific

Volume S=f [T,V]

We know that,

From Maxwell equation, We know that


Substituting in dS equations

multiplying by T

This is known as the second form of entropy equation (or) The second Tds equations.

(6) Show that the Joule-Thamson Co-efficient of an ideal gas is zero

[AU DEC 09,AU MAY O7]

The Joule-Thomson Co-efficient is defined as change in temp with change in pressure,

Keeping the enthalpy remains constant.

We know that the equation of state as

PV=RT

Differentiating the above equation of state with respect to ‘T’ by keeping pressure ‘P’

Constant

µ=0 it implies about the Joule-Thomson Co-efficient is Zero for ideal gas.


(7) Describe Joule Kelvin effect with the help of T-P diagram [AU MAY 07,08]

The Joule-kelvin effect (or) Joule Thomson effect is an efficient way of cooling gases. In

this, a gas is made to undergo a continuous throttling process.

The constant pressure is maintained at one side of porous plug and a constant lower

pressure at the other side. The apparatus lis thermally insulated so that heat loss can be

neglected.

Joule-Thomson co-efficient is defined as the change in temperature with change in

pressure keeping the enthalpy remains constant. It is denoted by (µ)

Throttling process :-

Throttling process is defined as the fluid Expansion through a minute orifice (or) slightly

Opened value.

During the throttling process, pressure and velocity are reduced.

But there is no heat transfer and no work done by the system.

In this process, enthalpy remains constant.

Joule-Thomson Experiment:-

In this experiment, a stream of gas at a pressure (p1) and temp (T1) is allowed to flow

continuously the porous plug.

The gas comes out from the other side of the porous plug at a pressure (p2) and

Temp (T2).

The whole apparatus is completely Insulated, Therefore, no heat transfer take places

Q=0

W=0 From steady flow energy equation,

We know that


Since there is no considerable change velocity V1=V2,Z1=Z2.

Q=0,W=0 and Z1=Z2 are applied in steady flow energy equation

So, the equation (1) becomes

h1=h2

Enthalpy at Inlet h1=Enthalpy at outlet h2

It indicates that the enthalpy is constant for throttling process.

It is assumed that a series of Experiment performed on a real gas keeping the Initial

pressure P1 and Temperature T1constant with various reduced down steam pressure

(P2,P3,P4….)

It is found that the down steam temp also changes

The result from these experiment can be plotted as a constant enthalpy curve on T-P

[Temp-Pressure] plane

The slope of a constant enthalpy is known as Joule-Thomson co-efficient it is denoted by

(µ).

For real gas (µ) may be either positive (or) negative depending upon the thermodynamic

state of the gas.


(8) A certain ideal gas has (R)=290 J/kg.K and =1.35

(a) Determine the values of a Cp and Cv

(b) The mass of the gas it is filled in a vessel of 0.5m3 capacity till the pressure inside

becomes 4 bar gauge and the temp is 270C

(c) If 40 kJ of heat is given to the vessel, when the vessel is closed.

Determine the resting temp and pressure take the atm presser=100 Kpa.

Given:-

(R)= J/kg.K=0.29 kJ/kg.K

( =1.35

Volume (V) =0.5m3

Pressure (P1)=4 bar*100=400 kJ/kg.K

Temp (T1) =270C+273=300K

Heat (Q) =40kJ

Pressure (P)atm=100 Kpa

To find:-

(1) Cp and Cv

(2) Mass of gas (m)

(3) Resulting temp (T2)

(4) Resulting pressure (P2)

Solution:-

Step;1 To find Cp andCv

R=Cp- Cv ---------------(1)

= ------------------(2)


1.35=

1.35 Cv=Cp ------------------(3)

R=Cp-Cv

290=1.35 Cv

828.5714 J/kg.K =CV ----------Sub in equation(3)

1.35 Cv=Cp

1.35*828.5714=Cp

CP=1118.5714 J/kg.K

Step:2 To mass of the gas(m)

P1V1=mRT1 =>gas equation

400*0.5=m*0.29*300

m=2.2988 kg

Step:3 To find Resulting Temp(T2)

Q=m*Cv*[T2-T1]

40=2.2988*0.8285[T2-300]

21.002=[T2-300]

21.002+300=T2

T2=321.0022 K

Step;4 To find Resulting Pressure [P2]


(9)A container of 3m3 capacity contains 10 kg of co2 at 270. Estimate the pressuree

Exereted by co2 by using

(1) Perfect gas equation

(2) Vander Waal’s equation

(3) Beattie Bridgeman equation [AU MAY 2012]

Given:-

(1)Volume of container (v) = 3m3

(2)Mass of co2 (mco2) = 10kg

(3)Temperature (T1) = 27oC + 273 = 300K

To Find:-

(1)Pressure Exerted by co2

Solution:-

Step: 1

Pco2 = mco2Rco2 T1

Where ,

Rco2 =

Rco2 =

Rco2 = 0.1889 kj/kg.K

Pco2 V = mco2 *Rco2*T1 Gas equation

Pco2

Pco2 =

2 = 188.9545 KN/m2


Step:2

To Find Pressure [Pco2]

Use Vander Waal’s gas

[P+ ] [V-b] = mRT

Where,

Pr Vr = 0.9 Note:- Pr = 3.375

Vr =

V = 0.267m3

Where,

Vc =

Vc = 11.23m3

PrVr = 0.9

Tc = Note:- Tr = 1.2

Tc =

Tc = 250 K

a =

a = 597

b =

b = *Vc

b =


[P + ] [V-b] = mRT

P = [ ]

P = [ ] – [ ]

P = -763.1224 – 66333

P = -829.4557 KN/m3

(10) One kg of ideal gas is heated from 50oC to 150oC. If R = 280 kj/kg.k and = 1.35

for the gas,

Determine,

(1) Cp and Cv

(2) Change in Internal energy

(3) Change in enthalpy

(4) Change in flow energy [Au May 2012]

Given:-

(1) Mass (m) = 1Kg

(2) Temperature (T1) = 50oC + 273 = 323K

(3) Temperature (T2) = 150oC + 27 3 = 423K

(4) = 1.35

(5) (R) = 280 kj/kg.K

To Find:-

(1) Cp and Cv

(2) (Δu) Change in internal energy

(3) (Δh) Change in enthalpy

(4) (Δpr)Change in flow energy


Solution:-

Step: (1)

Cv =

Cv = 875 kJ/kg.K

Cp = *Cv 1.32*87

Cp = 1155 KJ/kg.K

Step: (2)

Change in internal energy (Δu) = m* Cv [T2- T1]

(Δu) = 1*875[423 -323]

(Δu)= 8700J

Step:(3)

Change in enthalpy (Δh) = m*Cp*[T2-T1]

(Δh) = 1*1155[432-323]

(Δh) = 115500J

Step:(4)

Change in flow energy (Δpr) = Δh- Δu

= 115500 – 87500

(Δpr) = 2800J


(11) 0.4kg off co and 1kg of air is contained in a vessel of volume 0.43 at 15o C. Air has

23.3% of O2 and 76.7% of N2 by mass, Calculate the Partial pressure of each

constituent and total pressure in the vessel. Molar mass of Co, CO2 and N2 28,32 and 28

kg/k mol. [AU NOV 2005]

Given:-

(1) Mass of Co (mco) = 0.45kg

(2) Masss of air(mair) = 1kg

(3) Volume (V) = 0.4m3

(4) Temperature (T) = 15oC + 273 =288K

(5) Mole fraction of O2 (×o2) = = 0.233

(6) Mole fraction of N2 (×N2) = = 0.767

(7) Molar masses of co (MCo2) =28 kg/k mol

(8) Molar masses of Co2 (MCo2) = 32 kg/k mol

(9) Molar masses of N2 (MN2) 28 kg/k mol

To Find:-

(1) Partial Pressure of each constituent [Po2 , PN2 , Pco , Pa]

(2) Total Pressure in the vessel

Solution:-

Step:(1)

Gas constant of air (Ra) = [×o2× ] + [×N2 × ]

Ra = [0.233× ]

Ra = 0.0606 + 0.227

Ra = 0.288344 kj/kg.K


Step(2)

Mole fraction of [×co] =

[×co] =

[×co] = 0.3164

Step:(3)

Mole fraction of air (×air) = 1 - ×co

= 1 – 0.3164

( air) = 0.6835

Step:(4)

R = ×co + ×a

R = (0.3164 × ) + (0.6835 ×

R = 0.0939 + 0.0947

R = 0.18861 kj/kg.K

Step:(5)

Total Pressur (P) =

=

P = 196.9088 Kpa

Step:(6)

Partial Pressure of co(Pco) = ×co P

= 0.3164 × 196.9088


(Pco)= 62.3019 Kpa

Step:(7)

Partial Pressure of air (Pa) = ×air P

= 0.6835 × 196.9088

(Pair) = 134.5871 Kpa

Step:(8)

Partial Pressure of O2 (Po2) = ×o2 × Pair

Note:-

The air contains 23.3% O2 and 76.7% N2

(Po2) = ×o2 × Pair

(Po2) = 0.233 × 134.5871

(Po2) = 31.3588 Kpa

Partial Pressure of N2 (PN2) = ×N2 × Pair

= 0.767

(PN2) =103.22283 Kpa


(12) A certain gas has Cp = 0.913 kj/kg.K and Cv = 0. 653 kj/kg.K. Find the molecular

wight and specific gas constant (R) of the gas. [AU DEC 2006]

Given:-

(1) Cp =0.913 kj/kg.K

(2) Cv = 0.653 kj/kg.K

To find:-

(1) Molecular weight (M)

(2) Specific gas constant (R)

Solution:-

Gas constant (R) = Cp - Cv

= 0.913 – 0.653

(R)=0.26 kj/kg.K

R = 0.26 =

(M) = 31.98 kg/kg.mol.


13) A certain closed vessel has a capacity of 0.5 m3. It contains 20% nitrogen and

20% oxygen, 60% carbon-dioxide by volume at 20% and 1Mpa. Calculate the

molecular mass, gas constant, mass percentage and the mass of mixture.

[AU MAY 2008]

Given:-

Pressure (P) = 1Mpa ×10 → bar ×100 → 100 KN/m2

Temp (T) = 20°c + 273 → 293 K

Volume (V) = 0.5 m3

Mole fraction (MN2) = → 0.2

Mole fraction (Mo2) = → 0.2

Mole fraction (Mco2) = → 0.6

To find:-

1) Molecular mass

2) Gas constant (R)mix

3) Mass percentage

4) Mass of mixture (m)

Solution:-

Step: 1

Mass fraction O2 (X2) =

(X2) =

(X2) = 0.231


Mass fraction N2 (XN2) =

(XN2) =

Mass fraction co2 (Xco2) =

(Xco2) =

Step: 2

Partial pressure of O2 (Po2) = Xo2 × P

=0.231 × 10

Partial pressure of N2 (PN2) = XN2 × P

=0.264 × 10

Partial pressure of CO2 (PCO2) = XCO2 × P

=0.505 × 10

Step: 3 PV = mRT

m =

(XN2) = 0.264

(Xco2) = 0.505

(PO2) = 2.31 bar

(PN2) = 2.64 bar

(PCO2) = 5.05 bar


→ mass of oxygen =

=

→ mass of nitrogen =

=

→ mass of carbon-dioxide =

=

Step: 4

Mass of mixture = (m) = massO2 + mas sN2 + massCO2

=1.52 + 1.52 + 4.56

Step: 5

Equivalent gas constant of mixture (R)

(R) Mix =

=

Mass of oxygen = 1.52 Kg

Mass of carbon-dioxide = 4.56 Kg

(m) = 7.6 Kg

Mass of nitrogen = 1.52 Kg

(R) Mix = 0.225 KJ/Kg.k


14) A mixture of ideal gases consists of 2.5 Kg of N2 and 4.5 Kg of CO2 at pressure of

4 bar

and a temperature of 25°c. Determine,

(1) Mole fraction of each consistent

(2) Equivalent molecular weight of the mixture

(3) Equivalent gas constant of the mixture

(4) The partial pressure and partial volumes

(5) The volume and density of mixture [AU DEC 2009]

Given:-

Mass of N2 = 2.5 Kg

Mass of CO2 = 4.5 Kg

Pressure (P) = 4 bar × 100 → 400 KN/m2

Temperature = 25°c + 273 → 298 K

Solutions:-

Step: 1

To find mole fraction of N2

X =

X =

To find mole fraction of co2

Xco2 =

XN2 = 0.466


Xco2 =

Xco2 = 0.534


Step: 2

Equivalent molecular weight of the mixture

M Mix = XN2 MN2 + XCO2 Mco2

= (0.4666×28) + (0.534×44)

Step: 3

Equivalent gas constant of the mixture (R) Mix

(R mix) =

Where,

Total mass of mixture (m) Mix = mN2 + mco2

(m) Mix = 2.5 + 4.5

(m) Mix = 7 Kg

(R) Mix =

Step: 4

Partial pressure and partial volume

PN2 = XN2×P → 0.466×400 = 186.4 KN/m2

PN2 = Xco2×P→ 0.534×400 = 213.6 KN/m2

VN2 =

M Mix = 36.544 Kg/Kg mol

(R) Mix =0.288 Kg/Kg.k


=

Vco2 =

=

Step: 5

Density of mixture

Density of N2 =

=

=4.52 Kg/m3

Density of co2 =

=

= 7.103 Kg/m3

VN2 =0.553 m3

Vco2 = 0.6335 m3


(15) A tank contains 0.23 of gas mixture composed of 4kg of nitrogen, 1 kg of oxygen

and 0.5kg of carbon dioxide .if the temperature in 20oC, determine the total pressure, gas

constant and molar of the mixture. [AU MAY 2007]

Given:-

(1)Volume (v) = 0.2m3

(2)Mass of nitrogen (mN2) = 4kg

(3)Mass of oxygen (mo2) = 1 kg

(4)Mass of carbon dioxide(mco2) =0.5kg

(5) Temperature (T) = 20oC + 273=293K

To Find:-

(1) Total pressure

(2) Gas content (R)

(3) Molar mass of mixture (mmix)

Solution:-

Step:(1)

To Find gas constant of N2 O2,Co2

RN2 = = = 0.299 kj/kg.L

Ro2 = = = 0.2598 kj/kg.K

Rco2 = = 0.1889 kj/kg.K


Step:(2)

The partial pressure of each component

PN2 = =

PN2 = 1739.834 KN/m3

Po2 = =

Po2 =380.607 KN/m3

Pco2 = 138.3692 KN/m3

Step:(3)

Total pressure (P) = PN2 + PO2 + Pco2

(P) = 1739. 834 + 380.607 + 138.3693

(P) = 2258.8102 KN/m2

Step:(4)

The molar mass of mixture (m mix)


=0.1428+0.03125+0.01136

=0.1854 kg/kg.mole

=4+1+0.5

=5.5 kg

=29.6634 kg/k mole


(16) A closed rigid cylinder is divided by a diaphragm in to two equal compartments,

each of volume 0.1 m3. The pressure in one compartment is2.5 Mpa and in the other

compartments mixes to bring the pressure to a uniform value throughout the cylinder

which is insulated. Find the net change of entropy for the mixing process. [AU MAY 2012]

Given;-

Volume(VA=VB)=0.1m3

Pressure (PA) =2.5 Mpa×1000=2500 kN/m2

(PB) =1 Mpa×1000=1000 kN/m2

Temp (T)=200C+273=293 K

To find:-

Net change in entropy for mixing proess

Solution:-

Step:1

mass (mA)=?

PA VA= mART

Step:2

PB VB=mBRT


Step:3

The total mass of mixture=(mm)=mA+mB

(mmix)=2.9729+1.1891

(mmix)=4.162 kg

Step:4

Total Volume of mixture (Vm)=VA+VB

(Vm)=0.1+0.1

(Vm)=0.2 m3

Step:5

The total pressure of mixture (Pm)

Pm Vm=mmRT

Pm=1749.9337 kN/m2

Step:6

To find Net change in entropy ( Sm) of mixture

=2.9729×0.287 ln [ +1.1891×0.287 ln [

=-0.3043+0.1909

( S)mix=0.1134 kJ/kg . K


(17) An insulated rigid tank is divided into two compartment by a partition. One

compartment contains 7 kg of oxygen gas 400C and 100 Kpa and the other compartment

contains 4 kg of nitrogen gas at 200c and 150 Kpa. Now the partition is removed and the

two gases are allowed to mix. Determine

(1) The mixture temp

(2) The mixture pressure after equilibrium has been established. [AU DEC 10]

Given:-

Mass of oxygen(mo2)=7kg

Temp of oxygen (To2)=400C+273=313K

Pressure of oxygen(Po2)=100Kpa=100kN/m2

Mass of nitrogen(mN2)=4 kg

Temp of Nitrogen (TN2)=200C+273=293K

Pressure of Nitrogen (PN2)=150 Kpa=150kN/m2

To find:-

(1) Mixing temp (Tm)

(2) Mixing pressure (Pm)

Solution:-

Step:1

(Tm)=

(Tm)=

(Tm)=305.16K

Step:2

(Pm)=

Where

Mole of mixture (mm)=Xo2+XN2s


Mole of Oxygen=

Mole of nitrogen=

Mole of mixture (mm)=Xo2+XN2s

=0.219+0.142

(mm)=0.362 kg.mole

Where,

Volume of mixture (Vm)=

Volume of Oxygen (Vo2)=

(Vo2)=

(Vo2)=5.7 m3

Volume of nitrogen (VN2)=

=

(VN2)=8.02 m3

Volume of mixture (Vm)=

=5.7+2.32

(Vm)=8.02m3

Pressure of mixture (Pm)=

=

(Pm)=114.5 kN/m2

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