APPLICATIONS OF FIRST LAW OF THERMODYNAMICS
The first law of thermodynamics can be applied to some simple processes such as:
- Relation between and
- Boiling process iii) Isobaric process (no change in pressure) iv) Isochoric process (no change in volume)
v) Isothermal process (no temp change) vi)Adiabatic process (no heat learning or enters) RELATION BETWEEN (Mayer’s equation)
(Mayer’s equation) Consider n – moles of an ideal gas
Suppose the gas is heated at constant volume so that its temperature increases by The amount of heat Q supplied is given by
The amount of heat Q supplied is given by 
From the 1st law of thermodynamics
Q
The gas is heated at constant volume and hence
Equation (1) = Equation (2)
Suppose now that n – moles of the same gas are heated at constant pressure so that its temperature increases by the same amount T
T The heat Q‘ – supplied is given by
From
From ideal gas equation
Substitute equation (3), equation (4) and equation (6) in equation (5)
This equation shows that is greater than
Moreover;
- When the gas is heated at constant volume, no work is done ( V 0) and all the heat goes into raising the internal energy and thus the temperature of the gas.
- When the gas is heated at constant pressure, it expands and does work so that only a part of heat is used up in increasing the internal energy and hence temperature of the gas.
Therefore, in a constant- pressure process, more heat is needed to achieve a given temperature change than that of constant volume process.
NOTE:
If are specific heat capacities of the gas at constant pressure volume respectively then:
are specific heat capacities of the gas at constant pressure volume respectively then: 
Where r = gas constant for a unit mass
It is given by
TYPES OF GASES
- MONO-ATOMIC GAS
This is a type of a gas whose molecule consist of a single atom
Example
Helium
Neon
- DIATOMIC GAS
This is a type of gas whose, molecule consist of two atoms
Examples
Hydrogen (H2)
Oxygen (O2)
Chloride (Cl2)
POLYATOMIC GAS
This is a type of gas whose molecules consists of more than two atoms.
Examples
Carbon dioxide (CO2)
Ammonia (NH3)
Hydrogen peroxide (H2O2)
Steam (water gas) (H2O)
Etc.
DEGREES OF FREEDOM Degrees of freedom of a gas molecule are the number of independent ways the molecule can possess energy.
It refers to the axes x, y and z to which the gas molecule can move freely.
DEGREES OF FREEDOM FOR DIFFERENT GASES
MONOATOMIC GAS
The atom can move freely along x, y and Z- axes.
Hence a monatomic gas has three degrees of freedom
DIATOMCI GAS
The molecule has three degree of freedom in translation and two degree of freedom in rotation. Thus, a diatomic gas has a total 5 – degrees of freedom
POLYATOMIC GAS
The molecule has 3-degree of freedom in translation and 3-degrees of freedom in rotation.
Hence, a polyatomic gas has a total of 6-degrees of freedom.
EQUIPARTITION PRINCIPLE
The mean kinetic energy of a molecule is given by
= KT ……………………………………… (1) Where all symbols carry their usual meaning.
Since gas molecules are in random motion
………………………………….. (2) If is the resultant mean square speed of a gas molecule then: –
is the resultant mean square speed of a gas molecule then: – = + + ………………………….. (3) Substitute equation (2) in equation (3)
= = ………………………….. (4) Substitute equation (4) in equation (1)
= = = KT 
When are the main square speed of gas molecules along x, y and z-axes respectively, and the
axes of the degrees of freedom.
are the main square speed of gas molecules along x, y and z-axes respectively, and theaxes of the degrees of freedom.
Equation (5) above express the principal of equitation of energy which says: The mean energy of the molecules of a gas is equally divided among their available degrees of freedom, the average for each degree of freedom being KT
KT Where K Boltzmann Constant
Boltzmann Constant T Thermodynamic temperature
Thermodynamic temperature DEDUCTIONS FROM EQUATION (5)
- MONOATOMIC GAS
Has three degrees of freedom Mean K.E of a molecule KT 3
- DIATOMIC GAS
Has 5 – degrees of freedom
Mean K.E of a molecule = KT X 5
KT X 5 
- POLYATOMIC GAS
Has 6 – degrees of freedom
Mean K.E of a molecule = KT 6
RATIO OF MOLAR HEAT CAPACITIES OF A GAS
The ratio is called gamma
is called gamma 
This ratio has different values for different gases
FOR MONOATOMIC GAS
This means the internal energy U is given by:
U = KT
Where K = = Boltzmann
constant
= Boltzmannconstant
U = T
T 
If Cv is the molar heat capacity of a gas at constant volume then
U =
For 1 mole of a gas n = 1
U = 
…………………………………… (2) Equation (1) equation (2)
equation (2) 
………………….(3) According to Mayer‘s equation
– = R = R +
Substitute equation (3) in this equation:
Now = = = 1.67
= = = 1.67 ...
FOR DIATOMIC GAS
The internal energy U is given by:
U = KT
Where K
U = T
T 
The internal energy of the gas can also be given by
Equation (1) equation (2)
equation (2) RT =
According to Mayer‘s equation
– = R = R +
Substitute equation (3) in this equation
Now, = = = 1.40
= = = 1.40 
FOR POLYATOMIC GAS
The internal energy (U) of a gas is given
U 3KT
Where K
U 3 T
3 T The internal energy U of the gas can also be given by:
Equation (1) = Equation (2) | ||
3RT | = |  |
Cv = 3R ………………………………(3)
According to Mayer‘s equation
– = R = R +
Substitute equation (3) in this equation
+ 3R
Now, = = = 1.33
= = = 1.33 
APPLICATION OF =
= It is used to solve adiabatic problems
BOILING PROCESS
Suppose a liquid of mass m vaporizes at constant pressure P
Let be volume in the liquid state.
be volume in the liquid state. Let be volume in the vapor state
be volume in the vapor state Here expansion takes place at constant pressure and hence the work done by the system is
W =
) Where = Increase in internal energy of the system
= Increase in internal energy of the system MELTING PROCESS
When a solid changes into liquid state (melting), its internal energy increases.
This can be calculated from, the 1st law of thermodynamics.
Let m = mass of the solid
L = specific latent heat of fusion
Heat (Q) absorbed during the melting process is
Q = …………………………….. (1)
…………………………….. (1) Since during melting process, the change in volume ( = 0
= 0 From the 1st law of thermodynamics
Q = + W
+ W Q = +
+ = + P x 0 ... ……………………………(2)
……………………………(2) During melting process, internal energy increase by .
. Since temperature remains constant during melting the kinetic energy remains the same.
Therefore, the increase in potential energy.
ISOBARIC PROCESS
This is the process which occurs at constant pressure.
According to 1st law of thermodynamics
Q = +
+ In this case, the heat Q added increases the does internal energy of the gas as well as the gas does external work.
ISOCHORIC PROCESS
This is the process which occurs at constant volume (i.e = 0)
= 0) In such a process, external work done W is zero
W P 0 0
P 0 0 According to 1st law of thermodynamics:
Q = + W
+ W 
We conclude that if heat is added to a system at constant volume, all the heat goes into increasing the internal energy of the system.
ISOTHERMAL CHANGE /PROCESS
Definition
An isothermal change is that change which takes place at constant temperature.
In such a process heat is, if necessary, supplied or removed from the system at just the right rate to maintain constant Temperature.
Conditions for isothermal change
- The gas must be held in a thin-walled, highly conducting vessel, surrounded by a constant temperature bath.
- The expansion or contraction must take place slowly. So that the heat can pass in or out to maintain the temperature of the gas at every instant during expansion or contraction.
Isothermal change represented graphically
When the temperature is constant the pressure of a gas varies with volume and a graph which
shows this variation is a curve known as isothermal curve
Where
This graph is also called PV – curve or PV – Indicator diagram
When a gas expands, or is compressed, at constant temperature, its pressure and volume vary along the appropriate isothermal, and the gas is said to undergo an isothermal compression
expansion
Isothermal reversible change
When the gas is compressed isothermally from , T to , T then a graph which show this variation is:
, T to , T then a graph which show this variation is: 
If the gas is allowed to expand isothermally so that the state of the gas is brought back from
(, T) through exactly the same intermediate stage then the gas is said to undergo isothermal reversible change:
, T) through exactly the same intermediate stage then the gas is said to undergo isothermal reversible change: Definition
An isothermal reversible change is that change which goes to and from through exactly the same intermediate stages at constant temperature.
Isothermal reversible change equation
Since the temperature is constant, and is isothermal change obeys Boyle‘s law.
PV = constant.
... P1V1 = P2V2 Isothermal reversible change equation
Work done during isothermal change
Consider a gas pressure (P) expanded isothermally from volume , to volume .
, to volume . The gas does some work during the expansion given by:
= …………………….. (1) Where all symbols carry their usual meaning.
According to ideal gas equation
PV =
P = …………………………. (2)
…………………………. (2) Substitute equation (2) in equation (1)
dW = dV ……………………………….(3)
dV ……………………………….(3) When n, R and T are constants
The total work done W is obtained by integrating equation (3) above from volume , to volume .
, to volume . = W =
W
W =
The work done W can either be positive or negative depending on whether the gas undergoes isothermal expansion or compression.
If it is isothermal expansion then the work done is positive and the work is said to be done by the gas.
If it is isothermal compression then the work done is negative and the work is said to be done on the gas by the compressing agent.
The first law of thermodynamics applies to an isothermal change.
According to first law of thermodynamics
Q = +
+ Where,
Q Heat input
Heat input Increase in internal energy of the gas External work done For an isothermal change the temperature of the gas is constant and hence for an ideal gas there is no change in internal energy of the gas, since
or nR 
0 Hence Nr 0
Nr 0 0 Applying the first law of thermodynamics
Q = +
+ Q = 0 +
Q = External work done
External work done Which means the whole amount of heat energy supplied (Q) is used to perform external work
ADIABATIC CHANGE /PROCESS
An adiabatic change is the change which takes place without exchange of heat between inner and outer of the system.
It is the one which takes place at constant heat.
In general, an adiabatic change involves a fall or rise in temperature of the system.
Condition for adiabatic change
No heat is allowed to enter or leave the gas. therefore.
- The gas must be held in a thick – walled, badly, conducting vessel.
- The change in volume must take place rapidly to give as little time as possible for heat to escape.
Examples of adiabatic process/change
i. i. The rapid escape of air from a burst Tyre.
i. ii. The rapid expansions and contractions of air through which a sound wave is passing.
Adiabatic change represented graphically A curve which relates the pressure and volume when the heat content of the gas is kept constant is called an adiabatic.
Adiabatic curves and isothermal curves are similar except that adiabatic are steeper than isothermals.
If the gas is compressed adiabatically from volume to volume its temperature rises to T2 so that its new position is () on the new isothermal.
to volume its temperature rises to T2 so that its new position is (
) on the new isothermal. Similarly, if the gas is left to expand adiabatically from volume to volume its temperature is lowered to so that its new position is ( ) on the new isothermal Adiabatic reversible change
to volume its temperature is lowered to 
so that its new position is ( 
) on the new isothermal Adiabatic reversible change Definition
An adiabatic reversible change is the change which goes to and fro through exactly the same intermediate stages without exchange of heat between inner and out of the system.
Adiabatic reversible change equation
Consider an adiabatic change and the first law of thermodynamics
Q +
+ But – for 1 mole of a gas
– for 1 mole of a gas Since that is an adiabatic change Q = 0 i.e. no heat is allowed to enter or leave the system.
0 = + ………………………….. (1)
+ ………………………….. (1) From ideal gas equation for 1 mole
PV = RT
Differentiating this equation
We may write:
+ = = ………………………….. (2) Substitute equation (2) in equation (1)
= + = + ………………….. (3) According to Mayer‘s equation
R = – ………………………………………. (4) substitute equation (4) in equation (3)
– ………………………………………. (4) substitute equation (4) in equation (3) = = = V Dividing by throughout
throughout = + But = = ratio of molar heat
= = ratio of molar heat Capacities of a gas
+ We may write
0 =
Vd = –
= – = Integrating this equation
= = + K Where K = constant of integration
= K + = K 
= K 
= K = Where = constant
= constant 
In general
Adiabatic reversible change equation in terms of temperature.
From equation (5)
= constant = K ……………………………………….. (7) From ideal as equation
PV = RT
P = ………………………………………….. (8)
………………………………………….. (8) Substitute equation (8) in equation (7)
= K = But = constant
= constant 
In general
Work done during adiabatic process
Consider a gas which expands adiabatically such that its volume changes from The work done during the expansion is given by:
The work done during the expansion is given by: W = ……………………………………… (11)
……………………………………… (11) For an adiabatic change = K
= K P = 
W =
W = K
W = K
W =
W =
But, K = =
= W =
W =
W =
If the temperature of the initial state is and that of the final state is then ideal gas equation gives.
and that of the final state is 
then ideal gas equation gives. = nR and = nR
W =
The first law of thermodynamics applied to an adiabatic change / process
According to first law of thermodynamics
Q = +
+ Where all symbols carry their usual meaning.
For an adiabatic process there is no change in heat content of the gas i.e Q = 0
0 = +
+ = – Since = we have:
= we have: 
Here the work is done at the expense of the internal energy of the gas itself.
Thus, if the gas expands adiabatically, it does work and W = is positive, hence is negative, since the temperature of the gas decreases.
is positive, hence 
is negative, since the temperature of the gas decreases. If the gas is compressed adiabatically, the work is done on it and W = is negative, hence is positive, since the temperature of the gas increases.
is negative, hence is positive, since the temperature of the gas increases. From equation (13)
= = – = – But = W = work done during adiabatic process.
= W = work done during adiabatic process. W = –
W = –
W = – ………………………. (14)
………………………. (14) From ideal gas equation: PV = nRT
= = Equation (14) becomes:
W =
W =
W = …………………………….. (15)
…………………………….. (15) But = =
= = Equation (15) becomes:
W =
But =
= 
OR
Problem 63
5 moles of hydrogen initially at STP are compressed adiabatically so that the temperature becomes 400. Find:
. Find: - The work done on the gas
- The increase in internal energy of the gas
Given that = 1.4 for diatomic gas.
= 1.4 for diatomic gas. Problem 64
At 27 two moles of an ideal monatomic gas occupy a volume V. the gas expands adiabatically to a volume 2V. Calculate:
two moles of an ideal monatomic gas occupy a volume V. the gas expands adiabatically to a volume 2V. Calculate: - The final temperature of the gas
- The change in its internal energy
- The work done by the gas during this process
Given that = 1.67
= 1.67 Problem 65
A metallic cylinder contains 10 litres of air at 3 atmospheres of pressure and temperature of 300K.
- If the pressure is suddenly doubled, what are the new values of volume and temperature.
- If the pressure is slowly doubled, what are the new values of volume and temperature.
Problem 66
- Define the principle molar heat capacities of a gas.
- Why the energy needed to raise the temperature of a fixed mass of a gas by a specific amount is greater if the pressure is kept constant than when the volume is kept constant.
- Find the two principal heat capacities for oxygen (diatomic molecule) whose ratio of is 1.4 at STP.
Problem 67
A quantity of oxygen is compressed isothermally until its pressure is doubled. It is then allowed to expand adiabatically until its volume is restored. Find the final pressure in terms of the initial pressure. Given that = 1.4
= 1.4 Problem 68
- With the help of sketch diagram distinguish between an ―isothermal change‖ and an ―adiabatic change‖. Illustrate your answer with an example of a gas changing from state A to state B.
- Argon gas (specific heat capacity ratio 1.67) is contained in a 250 versel at a pressure of 750mmHg and a temperature of 0. The gas is expanded isothermally to a final volume of 400
- Calculate the final pressure of the gas
- By how much will the pressure will be lowered if the change is made adiabatically instead?
Problem 69
(a)
- Define mean ―free path‖ for a molecule of a gas
- How is the means free path of the molecule of a gas affected by temperature.
- The heat capacity at constant volume for 8 moles of oxygen gas is 166.2KJ-1 .
Find the heat capacity at constant pressure for 8 moles of oxygen.
Problem 70
- What is the difference between an ―isothermal process‖ and ―adiabatic process‖?
- How, much work is required to compress 5 moles of air at 20 and 1 atmosphere pressure at of the original volume by:
- An isothermal process
- An adiabatic process
- What are the final pressure for case (b) (i) and (b) (ii) above?
- In a diesel engine, the cylinder compresses air from approximately standard temperature and pressure to about one sixteenth of the original volume and a pressure of about 5 atmospheres. What is the temperature of the compressed air?
= 1.403 R = 8.31
= 20.68 
Problem 71
100g of a gas are enclosed in a cylinder which is fitted with a movable frictionless piston.
When a quantity of heat is supplied to the gas it expands at constant pressure doing 8400J of work and heating up by 20. Calculate:
. Calculate: - The change in internal energy of the gas
- The specific heat capacity of the gas at constant volume
Given that = 1.26 x
= 1.26 x 
Problem 72
Given that the molar heat capacities of hydrogen at constant value and constant pressure are respectively 20.5 and 28.8.
and 28.8. Calculate
- The molar gas constant
- The heat needed to raise the temperature of 8g of hydrogen from 10 to 15 at constant pressure.
- The increase in internal energy of the gas
- The external work done
Problem 73
The density of a gas is 1.775 and 1 pressure and its specific heat capacity at constant pressure is 846 . Find the ratio of its specific heat capacity at constant pressure to that at constant volume.
and 1 pressure and its specific heat capacity at constant pressure is 846 . Find the ratio of its specific heat capacity at constant pressure to that at constant volume. Problem 74
A gas of volume 500 and pressure 1.0 x Nm expands adiabatically to
and pressure 1.0 x 
Nm expands adiabatically to 600. Calculate
. Calculate - The find pressure
- The work done by the gas
- The final temperature if the initial temperature of the gas before expansion was 23
Given that
Problem 75
One gram of water becomes 1671 of steam at a pressure of 1 atmosphere. The latent heat of vaporization at this pressure is 2256 . Calculate the external work done and the increase in internal energy.
Problem 76
1.0 of water is converted into 1671 of steam at atmospheric pressure and 100 temperature. The latent heat of vaporization of water is 2.3
of steam at atmospheric pressure and 100 temperature. The latent heat of vaporization of water is 2.3 . If 2.0Kg of water is converted into steam at atmospheric pressure and 100 temperature, then how much will be the increase in its internal energy?
Given that
Density of water = 1.0
Atmospheric pressure = 1.01
Problem 77
- What is an isothermal change?
- A cylinder fitted with a frictionless piston holds a volume of 1000 of air at a pressure of 1.10 Pa and temperature of 300 K. The air is then heated to 375K at constant pressure. Determine the new volume of the gas.
The gas is then compressed isothermally to a volume of 1000. Calculate the new pressure.
. Calculate the new pressure. Problem 78
- (i) What is the difference between an isothermal and an adiabatic process?
- Show that an adiabatic change follows an adiabatic equation.
= constant - (i) Distinguish between the specific heat capacity and the molar heat capacity. Given the unit of each.
- Calculate the two principal molar heat capacities of oxygen and explain why the specific heat capacity of the gas at constant volume is less than that at constant pressure.
Problem 79
i. ( ) What is reversible change? ii. ( ) State the condition for a reversible change to occur.
iii. ( )A litre of air at Pa pressure expands adiabatically and reversibly to twice its volume.
Pa pressure expands adiabatically and reversibly to twice its volume. Calculate the work done by the gas.
Problem 80
A cylinder in the figure below holds a volume = 1000 of air at an initial pressure of = 1.1 Pa and temperature = 300K. Assume the air behaves like an ideal gas.
= 1000
of air at an initial pressure of = 1.1 Pa and temperature = 300K. Assume the air behaves like an ideal gas. 
i. (i ) AB – the air heated to 373 K at constant pressure. Calculate the new volume.
i. ( ii) BC – the air is compressed isothermally to volume
i. (iii) Calculate the root mean square speed of nitrogen molecules at a temperature of
27
Problem 81
- ( ) State the 1st law of thermodynamics and write its equation.
- ( ) A liter of air initially at 25 and 760mmHg is heated at constant pressure until the volume is doubled. Determine:
- The final temperature
- The external work done by the air in expanding it. (iii) The quantity of heat supplied
Problem 82
0.15 mol of an ideal mono atomic gas is enclosed in a cylinder at a pressure of 250 KPa and a temperature of 320K. The gas is allowed to expand adiabatically and reversibly until its pressures is 100KPa
- Sketch a P – V curve for the process.
- Calculate the final temperature and the amount of work done by the gas.
Problem 83
i. ( ) Define the bulk modulus of a gas
i. ( i) Find the ratio of the adiabatic bulk modulus of a gas to that of its isothermal bulk modulus in terms of the specific heat capacities of the gas.
Problem 84
- A gas expands adiabatically and its temperature falls while the same gas when compressed adiabatically its temperature rises. Explain giving reasons why this happens.
- A mole of oxygen at 280K is insulated in an infinitely flexible container is 5
. When 580J of heat is supplied to the oxygen the temperature increases to 300K and the volume of the container increases by 3.32 . Calculate the values of the principal molar heat capacities and the specific universal gas constant.
. Calculate the values of the principal molar heat capacities and the specific universal gas constant. Given that molar mass of oxygen = 32
Problem85
(a) (i) Why is heat needed to change liquid water into vapour?
What amount of energy is needed
(ii) The molar heat capacity of hydrogen at constant volume is 20.2
What is the molar heat capacity at constant pressure?
(b) In an industrial refrigerator ammonia is vaporized in the cooling unit to produce a low temperature. Why should the evaporation of ammonia reduce the temperature in the refrigerator?
How much energy is needed to convert 150g of water at 20 into steam at
into steam at 100
Problem 86
An ideal gas is kept in thermal contact with a very large body of constant temperature T and undergoes an isothermal expansion in which its volume changes from .
. Derive an equation for the work done by the gas.
Problem 87
A heat engine carries I mole of an ideal gas around a cycle as shown in the figure below. Process 1 – 2 is at constant volume, process 2 – 3 is adiabatic and process 3 – 1 is at a constant pressure of 1 a.t.m. The value of for this gas is .
for this gas is 
. Find:
i. ) The pressure and volume at points 1, 2 and 3 ii ) The net work done by the gas in the cycle.
Problem 88
The door of a working refrigerator is left open.
What effect will this have on the temperature of the room in which the refrigerator is kept?
Explain
Problem 89
- a) What do you understand by the terms:
- Critical temperature?
- Adiabatic change?
- b) An air bubble is observed in a pipe of the braking system of a car. The pipe is filled with an incompressible liquid (see figure below). When the brakes are applied, the increased pressure in the pipe causes the bubble to become smaller.
Before the brakes are applied the pressure is 110 , the temperature is 290K and the length of the bubble is 15mm. When the brakes are applied quickly, the air bubble is compressed adiabatically and if the change in its length exceed 12mm the brakes fail. If the internal cross-sectional area of the pipe is 2
, the temperature is 290K and the length of the bubble is 15mm. When the brakes are applied quickly, the air bubble is compressed adiabatically and if the change in its length exceed 12mm the brakes fail. If the internal cross-sectional area of the pipe is 2 - Explain briefly why the compression of the bubble is considered to be adiabatic.
- What is the maximum safe pressure in the system during rapid braking if the bubbles change in length does not exceed 12mm? Take = 1.4
- Determine the temperature of the air in the bubble at the end of the adiabatic compression.
Problem 90
- Find the number of molecules and their mean kinetic energy for a cylinder of volume 4 containing oxygen at a pressure of 2 Pa and a
temperature of 300K
- When the gas is compressed adiabatically to a volume of 2, the temperature rises to 434K. Determine the , the ratio of the principal heat capacities.
Given that:
Molar gas constant = 8.31
And
Problem 91
- The first law of thermodynamics is a consequence of the law of conservation of energy. Explain briefly.
- What is the difference between isochoric process and isobaric process?
- Why is the energy needed to raise the temperature of a fixed mass of a gas by a specific amount is greater if the pressure is kept constant than if the volume is kept constant?
A certain volume of a dry air at S.T.P is allowed to expand four times its original volumes under:
- Isothermal conditions
- Adiabatic conditions
Calculate the final pressure and temperature in each case
Given that = 1.4
= 1.4 Problem 92
In a P – V diagram shown above, an adiabatic and an isothermal curve for an ideal gas intersect. Show that the absolute value of the slope of the adiabatic is
is times that of the isothermal Hence the adiabatic curve is steeper because the specific heat ratio is greater than 1
is greater than 1 Problem 93
A Tyre has air pumped at a pressure of 4 atmospheres at room temperature of 27. If the Tyre bust suddenly, calculate the final temperature (take = 1.4)
. If the Tyre bust suddenly, calculate the final temperature (take = 1.4) Problem 94
Two moles of oxygen are initially at a temperature of 27 and volume of 20 litres. The gas expanded first at constant pressure until the volume has doubled, and then adiabatically until the temperature returns to the original value.
and volume of 20 litres. The gas expanded first at constant pressure until the volume has doubled, and then adiabatically until the temperature returns to the original value. - What is the total increase in internal energy?
- What is the final volume?
Given that = = 2
= = 2 Problem 95
The Specific heat capacity of hydrogen at constant volume is 1.01. If the density of hydrogen at S.T.P is 0.09, calculate the specific heat capacity of hydrogen at constant pressure.
. If the density of hydrogen at S.T.P is 0.09
, calculate the specific heat capacity of hydrogen at constant pressure. Problem 96
(i) Does a gas do work when is expands adiabatically? If so what is the course of energy needed to do this work.
(ii)Derive a relation between the bulk modulus K and density ρ of a perfect gas under isothermal conditions and adiabatic conditions.
A mass of air at 27 and 750mmHg pressure occupies a volume of 8litres. If the air expands first isothermally until its volume increases by 50% and then adiabatically until its volume again increases by 50% each time reversibly. Calculate
and 750mmHg pressure occupies a volume of 8litres. If the air expands first isothermally until its volume increases by 50% and then adiabatically until its volume again increases by 50% each time reversibly. Calculate - The final pressure
- The final temperature
An ideal gas expands adiabatically from initial temperatures to a final temperature , prove that the work done by the gas is
to a final temperature , prove that the work done by the gas is Problem 97
An ideal gas at 760mmHg is compressed isothermally until its volume is reduced to 75% of its original volume. The gas is then allowed to expand adiabatically to a volume 120% of its original volume. If the temperature of the gas is 20
- Construct the P – V indicator diagram.
- Calculate the final pressure and temperature
Given that:
= 3600 J/kg K = 2400 J/kg K
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