Perfect gas :- A perfect gas or an ideal gas may be define as a stable of substance whose evaporation from its liquid state is complete known as perfect or ideal gas.
- Boyle’s Law
- Charle’s Law
- Gay Lussac Law
- General Gas Equation
- Avagadro’s Law
- Universal Gas Constant
- Regnault’s Law
- Relationshhip between specific heat and Cp and Cv
- Triple Points Of Water
- Thermal Expansion
- Question And Answer
This law was founded by Rovest Boyle in 1662. This law state the absolute pressure at given mass of perfect gas varies in inversely as its volume provide the constant temperature.
P ∝ 1/V
PV = constant
The more useful form of the above equation
P1V1 = P2V2 = P3V3 ……………… = constant.
This law was founded by French mam AC charle’s in 1787. This law may be state in the following two ways :-
A)The volume is given mass of a perfect gas varies directly proportional to absolute temperature provided the pressure remain constant
V ∝ T
V/ T = constant
V1/T1 = V2/T2 = V3/T3 ………… = constant.
All perfect gas change in volume by1/273th of its original volume of 0oC far every 1oC change in temperature , when the pressure remain constant.
B) At a given volume the pressure of a given mass of a perfect gas is directly proportional to its absolute temperature.
Vt = V0 + 1/273 V0 * t
Vt = V0 [273+t/273]
Vt = V0/T0
Gay lussac law
The absolute pressure of a perfect gas is directly proportional to its absolute temperature when its volume remains constant.
P ∝ T
P/T = constant
P1/T1 = P2/T2 = P3/T3 ………… constant.
General gas equation
It is derived by consider the fact that all the free variable i.e, pressure , volume , and temperature change simutanously.
General gas equation (Ideal gas equation)
PV = nRT
Boyle’s law = V ∝ 1/P
Charle’s law = V ∝ T
Avogadro hypotheses :- V ∝ n
V ∝ nT/P
VP ∝ nT
PV = nRT
This law state equal volume of all gasses will have equal number of molecules under identical condition of temperature and pressure remain constant.
Volume of gass . number of molecules
V ∝ n (constant temperature and pressure)
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Universal gas constant
The universal gas constant or molar constant of a gasis the product of the gas constant and the molecular of gas
Ru = MR
Where m = molecular mass of gas in kg mole.
R = gas constant
Ru = M1R1 = M1R1 …..
This law state the specific heat Cp (specific heat at constant pressure) and Cv (specific heat at constant volume) of a gas do not change with the change of temperature and pressure.
Specific heat :– The amount of heat required to raise the temperature of its unit mass through 1o
Specific heat at constant pressure :- It is the amount of heat required to raise the temperature of a unit mass of a gas through 1o , when its pressure is kept constant.
Qp = mCp ( T2 – T1)
Specific heat at constant volume :–
It is the amount of heat required to raised the temperature of a unit mass of a gas through 1o , when its volume kept constant.
Qv = mCv ( T2 – T1)
SI unit = 0.72kj/kg
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Relationshhip between specific heat and Cp and Cv
Let us consider in kg of a gas enclosed in a container and is being heated at constant pressure.
T1 = initial temperature of the gas
T2 = final temperature of the gas
V1 = intial volume of the gas
V2 = final volume of the gas
Cp = specific heat at constant pressure
Cv = specific heat at constant volume
P = absolute constant pressure
We know that heat supplied to the gas at constant pressure.
Q = mCp ( T2 – T1) ………………………………… 1
We also know that a part of heat is utilized in doing external work , while the remaining part of heat is used in increasing the internal energy of gas.
Q = w + du ……………………………………………… 2
Now heat utilized for external work.
W = P (V2 – V1) ……………………………………….. 3
And increasing in internal energy .
Dv = mCv (T2 – T1) …………………………………….. 4
Putting the value
of w equation dv in equation 2
Q = P (V2 – V1) + mCv (T2 – T1) …………………………………. 5
Using characterstic gas equation.
We have ,
For intial condition
Pv1 = MRT1
Pv2 = MRT2 …………… for final compare
P (V2 – V1) = MR (T2 – T1) …………………………………… 6
Putting the equation of P (V2 – V1) from eqution 6 in equation 5
Now we get ,
Q = MR (T2 – T1) + mCv (T2 – T1)
Now putting the value of Q in equation 1
mCp (T2 – T1) = MR (T2 – T1) + mCv (T2 – T1)
in absence equation we get ,
Cp = R + Cv
Cp – Cv = R
The equation may be written as
Cv (Cp/Cv – 1) = R
Cv = R/v – 1
The value of R is taken as 287 J/kg or 0.287 wJ/kg
The equation may be also written
Cp = Cv + R
Dividing both side by Cv we get
Cv/Cp = 1 + R/Cv
3m3 of a gas at 7 bar abs pressure expands to a volume of 12m3 at constant temperature. Find the final pressure of the gas
Ans.Given :- V1 = 3m3 , P1 = 7 bar , V2 = 12m3
Temperature = constant
Let P2 is the final pressure of the gas
We know that
P1V1 = P2V2
P2 = p1V1/V2 = 7*3/12
= 1.75 bar abs.
Determine the volume occupied by a given mass of air at a temperature of 245oC , if the same mass of air occupies 3m3 at 32oC . the pressure of the air remains unchanged.
Ans.Given :- T1 = 23oC = 296 k , T2 = 245oC = 578 k
V1 = 2m3 , pressure = constant
Let V2 is the volume occupied by air at 245oC
We know that ,
V2 = V1T2/T1 = 2*518/296
V2 = 3.5m3
A certain gas occupies 2m3 at a temperature of 127oC. the pressure of the gas is 7 bar and the gas expand in such as manner that the volume becomes 8m3 and the temperature 15oC. what will be the pressure of gas after expansion ?
Ans.Given :- V1 = 2m3 , T1 = 127o C = 400 k
P1 = 7 bar , V2 = 8m3 , T2 = 15o C = 288 k
Let P2 is the pressure of gas after expansion.
Now we know that ,
P1V1/T1 = P2V2/V2 (gas equation)
P2 = P1V1T2/V2T1
P2 = 1.26 bar.
Triple point of water
The triple point of water is the state at which the three phase of water namely ice , liquid , water vapour are equally stable and co-exist in equilibrium. It is unique because it occurs at a specific temperature of 273.16k and a specific pressure of 0.46 cm of Hg column. Thus for water ,
Ptr = 273.16k or 0.01oc
Almost all solids , liquids and gases expand on heating. The increase in the size of a body when it is heated is called thermal expansion.
Different typres of thermal expansion :-
- Linear expansion :- It is the increase in the length of a metal rod on heating.
- Superficial expansion :- It is the increase in the surface area of a metal sheet on heating.
- Cubical expansion :- It is the increase in the volume of block on heating.
Ans. Cause of thermal expansion :- All solids consist of atoms and molecules. At any given temperature , these atoms and molecules are held at equilibrium distance by force of attraction.
When a solid is heated , the amplitude of vibration of its atoms and molecules increases. The average inter-atomic separation increases. The results in the thermal expansion of the solid.
Ans. Latent heat :- When a solid change into liquid or a liquid changes into gas , it absorbs heat. But this heat does not show up an increase in temperature. This heat , used to change the state , is hidden or latent and is therefore called latent heat.
Ans. Water equivalent :- The water equivalent of a body is define as the mass of water which requires the same amount of heat as is required by the given body for the same rise of temperature.
Water equivalent = mass * specific heat
Or w = mc
The CGS unit of water equivalent is g and the SI unit is Kg.
Ans. Melting point :- The temperature at which the solid and the liquid states of a substance co-exist in thermal equilibrium with each other is called its melting point.
Boiling point :– the temperature at which the liquid and vapour states of a substance co-exist in thermal equilibrium with each other is called its boiling point.
Ans. Ideal gas law :- The ideal gas law states that the pressure, temperature, and volume of gas are related to each other.
Ans. Development of the ideal gas law :-The pressure, volume, temperature, and amount of an ideal gas are related by one equation that was derived through the experimental work of several individuals, especially Robert Boyle, Jacques A.C. Charles, and Joseph Gay Lussac.
Ans. The ideal gas law relates the four independent physical properties of a gas at any time. The ideal gas can be used in Stoichiometry problems in which chemical reaction involve gases.