01 part2-ideal-gas-problems-01

Ideal Gas

S.GunabalanAssociate ProfessorMechanical Engineering DepartmentBharathiyar College of Engineering & TechnologyKaraikal - 609 609.e-Mail : gunabalans@yahoo.com

Part - 2

Ideal Gas Equation of stateFor perfect gas, the ideal gas equation is

( Molar Volume )

Ideal Gas Equation of state

mas of gas in Kg

Ideal Gas Equation of state

( Molar Volume )

= = m3 = Volume (V)

Ideal Gas Equation of stateFor perfect gas, the ideal gas equation is ( Molar Volume )

= = m3 = Volume (V) Specific Volume –(m3 /Kg)Density - Kg/ m3 Molar Volume

specific Volume V – Volume m3

Ideal Gas Equation of state ( Molar Volume )

= = m3 = Volume (V)

Molar Volume specific Volume V – Volume m3

Units of Pressure

• atmospheric pressure (1 Atm.)= 1.013 bar, = 101325 Pascal (Pa = N/m²); = 760 millimeters of mercury absolute

(mmHgA) = 760 Torr (1 Torr = 1 mm HgA)

Gas Constant RGas Molar Weight ( M)Kg/Kmol Gas Constant (R )KJ/KgK

Air 28.97 0.287

Nitrogen 28.01 0.297

Oxygen 32 0.260

Hydrogen 2.016 4.124

Helium 4.004 2.077

Carbon dioxide 44.01 0.189

Steam 18.02 0.461

Note:

The specific volume of a substance is the ratio of the substance's volume to its mass. It is the reciprocal of density and is an intrinsic property of matter.

Substance Name Density Specific Volume

Kg/m3 m3/Kg

Air 1.2 0.83

Ice 916.7 0.00109

Water (liquid) 1000 0.00100

Salt Water 1030 0.00097

Mercury 13546 0.00007

Note: Molar volume

• The molar volume is the volume occupied by one mole of a substance at a given temperature and pressure.

• It is equal to the molar mass (M) divided by the mass density (ρ).

Specific heat

• The specific heat - the amount of heat required to raise a unit mass of the substance through a unit rise in temperature.

• For gases– At constant pressure

• Cp

– At constant Volume• Cv

This classic relationship between the specific heats of an ideal gas is called Mayer’s equation

Specific heat

• The specific heat - the amount of heat required to raise a unit mass of the substance through a unit rise in temperature.

• The product of mass and specific heat (mCv) is called the heat capacity(Cv) at constant volume (J/K)

• The latent heat is the amount of heat transfer required to cause a phase change in unit mass of a substance at a constant pressure and temperature

Questions -1

P = ?

V = Givenm = > M (Molar Weight) x one K Mole of N2

= 28.01 = m

R

• specific volume ( of the gas

Gas Molar Weight ( M)Kg/Kmol

Air 28.97

Nitrogen 28.01

= 1.4 given – (1)

Mayer’s equation

• – Two Equation two Variables

– R – is known from part a).

R

c)

Vessel capacity is same so V1 = V2

specific internal energy (kJ/kg) =specific enthalpy (kJ/kg) =Increase in specific entropy (kJ/kgK) =Magnitude and sign Heat transfer (kJ) = + w

= m + p(V2-V1)

• Given data• Vessel capacity (Volume)

– V = 3 m3

• Contain 1 K Mole of N2 At T = 90 0C

Answera) Calculate the pressure () and specific volume

( of the gas

From above table one K Mole of N2 = 28.01 = m

R 0.2968 KJ/KgKSubstituting in the ideal gas equation

Kelvin

945.58959 KN/m2

Gas Molar Weight ( M)Kg/Kmol

Air 28.97

Nitrogen 28.01

J = Nm

and

• specific volume ( of the gas

b) If the ratio of specific heat is 1.4, evaluate value of Cp and Cv

• ratio of specific heat is 1.4

Mayer’s equation

• = 1.4

• 0.297 KJ/KgK • 0.297 • = 0.7425 KJ/KgK• = 1.4 • = 1.4 • = 1.0395 KJ/KgK

C) Subsequently, the gas cools to atmospheric temperature of 20 0C; evaluate the final pressure of the gas.T2 = 20 0C

P2 = ?

Vessel capacity is same so V1 = V2

Substituting the values

= 763.2445

From Given dataVessel capacity (Volume) V1 = 3 m3

T1 = 90 0CCorresponding calculated 945.58959 KN/m2

d) Evaluate the increase in specific internal energy, Enthalpy , the increase in specific entropy and magnitude and sign of heat transfer.specific internal energy (kJ/kg) = = 0.7425 KJ/KgK

= -51.975 KJ/Kgspecific enthalpy (kJ/kg) = = 1.0395 KJ/KgK

= -72.765 KJ/Kg

T1 = 90 0CT2 = 20 0C = 0.7425 KJ/KgK = 1.0395 KJ/KgK

Increase in specific entropy (kJ/kgK) =Constant volume V1 = V2

= = = - 0.06908kJ/kgK

T1 = 90 0CT2 = 20 0C = 0.7425 KJ/KgK = 1.0395 KJ/KgK

Magnitude and sign Heat transfer (kJ) = + w

= m + p(V2-V1)

Constant volume V1 = V2

p(V2-V1) = 0

= 28.01 Kg x 0.7425 KJ/KgK

= 28.01 Kg x -51.975 KJ/Kg = - 1455.82 KJWhen work is done by a system, it is arbitrarily taken to be positive, andwhen work is done on a system, taken to be negative

Resulta)• 945.58959 KN/m2

b) = 0.7425 KJ/KgK = 1.0395 KJ/KgKc) = 763.2445 d)

= -51.975 KJ/Kg = 72.765 KJ/Kg = - 0.06908kJ/kgK = -1455.82 KJ

Exercise-1

• 1 Kg mole of N2 is contained in a vessel of volume 2.5 m3 at 100 . – i) find the mas, the pressure and specific volume of the

gas.– ii) if the ratio of the specific heat is 1.4, evaluate the

value of Cp and Cv– iii) Subsequently, the gas cools to atmospheric

temperature of 30 , evaluate the final pressure of the gas– iv) Evaluate the specific internal energy, the increase in

specific enthalpy, increase in specific entropy and sign of heat transfer

Question-2

A cylinder of 60 liter capacity contains CO2 at 10 bar and 20 . Determine mas of the gas, molar volume, density and specific volume of the gas.(Nov 2011)

V = 60 liters of CO2

P = 10 barT = 20 °C

DetermineMass of the gas (m)Molar Volume ()Density ()Specific volume ()

Molar Volume specific Volume V – Volume m3

Procedure

DetermineMass of the gas (m)

Molar Volume ()

Molar Volume

Molar Volume specific Volume V – Volume m3

Procedure

DetermineDensity ()

Specific volume ()Molar Volume specific Volume V – Volume m3

Ans:Mass of the gas (m)

P V⤫ m⤫ R T

Given dataV = 60 liters of CO2

P = 10 barT = 20 °C

atmospheric pressure (1 Atm.)1 Atm = 1.013 bar, 1 Atm = 101325 Pascal (Pa = N/m²);

Unit conversion1 bar = 1x 105 N/m2

1 m3 = 1000 litersJ = Nm

DetermineMass of the gas (m)

P = 10 bar V = 60 liters ⤫ m⤫ R T = 20 °C

Given dataV = 60 liters of CO2

P = 10 barT = 20 °C

Gas Molar Weight ( M)Kg/Kmol

Gas Constant (R) KJ/KgK

Air 28.97 0.287

Nitrogen 28.01 0.297

Oxygen 32 0.260

Hydrogen 2.016 4.124

Helium 4.004 2.077

Carbon dioxide 44.01 0.189

Steam 18.02 0.461

atmospheric pressure (1 Atm.)1 Atm = 1.013 bar, 1 Atm = 101325 Pascal (Pa = N/m²);

Unit conversion1 bar = 1x 105 N/m2

1 m3 = 1000 litersJ = Nm

DetermineMass of the gas (m)

P = 10 bar = 10 x 105 N/m2

V = 60 liters = 60/1000 m3

⤫ M = find⤫ R = cal for the substitution T = 20 °C = 20 +273 K

Gas Molar Weight ( M)Kg/Kmol

Gas Constant (R) KJ/KgK

Air 28.97 0.287

Nitrogen 28.01 0.297

Oxygen 32 0.260

Hydrogen 2.016 4.124

Helium 4.004 2.077

Carbon dioxide 44.01 0.189

Steam 18.02 0.461

atmospheric pressure (1 Atm.)1 Atm = 1.013 bar, 1 Atm = 101325 Pascal (Pa = N/m²);

Unit conversion1 bar = 1x 105 N/m2

1 m3 = 1000 litersJ = Nm

⤫ R = cal for the substitution for Carbon dioxideR KJ/KgKMass of the gas (m)

10 10 5 N/m2 m3 = 10 10 5 N/m2 m3 = ---------- kg

Gas Molar Weight ( M)Kg/Kmol

Gas Constant (R) KJ/KgK

Air 28.97 0.287

Helium 4.004 2.077

Carbon dioxide 44.01 0.189

Steam 18.02 0.461

P = 10 bar = 10 x 105 N/m2

V = 60 liters = 60/1000 m3

⤫M = find⤫R = cal for the substitutionT = 20 °C = 20 +273 K

DetermineSpecific volume ()

Density ()

Given dataV = 60 liters of CO2

P = 10 barT = 20 °C

DetermineMolar Volume

10 10 5 N/m2 (20+273) K = 0.410488

Given dataV = 60 liters of CO2

P = 10 barT = 20 °C

Result

= 0.410488

Exercise -2

Find the molecular weight and gas constant for the gas, whose specific heats are as followsCp = 1.967 KJ/Kg KCv = 1.507 KJ/Kg K

Exercise -2

– Mayer’s equation

Reference• Rajput, R. K. 2010. Engineering thermodynamics. Jones and Bartlett

Publishers, Sudbury, Mass.• Singh, O. 2003. Applied thermodynamics. New Age International (P) Ltd.,

Publishers, New Delhi.• Nag, P. K. 2002. Basic and applied thermodynamics. Tata McGraw-Hill,

New Delhi.