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TRANSCRIPT
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Nishant Gupta D-122 Prashant vihar Rohini Delhi-85
MEASURABLE PROPERTIES OF GASES1. MASS
The amount of gas is expressed in terms of its number of moles. For a gas of molar mass (M), the
mass in gram (w) is related to the number of moles (n) as n=w/M
2. VOLUME
The volume of the gas is the space occupied by its molecules under given set of conditions.
Volume of the container in which the gas is enclosed. 1m3 = 103 L = 103 dm3 = 106 cm3
3. PRESSURE
The force experienced by the walls of container due to bombardment of gas molecules . This
force per unit area of the walls is termed as gas pressure.
SI unit of force is kg ms-2 which is called Newton (N). I N= 1 kg ms-2
SI unit of pressure as N/m2 or Nm-2. This SI unit of pressure is called Pascal (Pa).- Kg m -1 S-2
I Pa=1 N m
-2
I bar = 100 kPa = 10
5
Pa
ATMOSPHERE AND ATMOSPHERIC PRESSUREA thick blanket of air which surrounds the earth is called atmosphere.
Molecules of various gases present in the atmosphere are under constant pull of the gravitational
force of the earth. As a result of this, atmosphere is dense near the surface of the earth than that at
high altitudes. Force experienced by any area of the earth exposed to the atmosphere is equal to
the weight of column of air above it. This force per unit area of the earth is called atmospheric
pressure.
1 atm = 76.0 cm of mercury =760 mm of mercury = 760 torr. = 1.01325 x 105 Pa
MEASUREMENT OF GAS PRESSUREThe pressure other than atmospheric pressure are measured by device is called manometer.
There are two types of manometer open end and closed end manometers
(a)Open end manometer. It consists of U tube partially filled with mercury. One limb of thetube is shorter than the other. The shorter limb is connected to the vessel containing the gas
whereas the longer limb is open .The mercury in the longer tube is subjected to the
atmospheric pressure while mercury in the shorter tube is subjected to the pressure of the gas.
MEASUREMENT OF GAS PRESSURE.There are three possibilities described as follows :
(i) If the level of Hg in the two limbs is same, then, gas pressure (Pgas)=atmospheric pressure
(Patm)
GASEOUS STATE - I
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(ii)If the level of Hg in the longer limb is higher , Gas pressure (Pgas) =Patm + (differencebetween the two levels)
= Patm + Ph
(iii)If the level of Hg in the shorter limb is higher, then Gas pressure (Pgas) =Patm-(difference
between the two levels)= Patm - Ph.
(b)Closed end manometer.This is generally used to measure low gas pressures. It alsoconsists of U-tube with one limb shorter than the other and partially filled with mercury.
The space above mercury on the closed end is completely evacuated. The shorter limb is
connected to the vessel containing gas. The gas exerts pressure on the mercury in the
shorter limb and forces its level down.
Gas pressure (gas) =[Difference in the Hg level in two limbs] Or Pgas = Ph
Closed end manometer.
(4)TEMPERATURETemperature may be defined degree of hotness. Common temperature-measuring deviceis thermometer.
(5)STANDARD TEMPERATURE AND PRESSURE (ST.P.)S.T.P. conditions (O0 or 273.15 K temperature and 1 atm (=1.01325 bar) pressure.
Standard Ambient Temperature and Pressure (SATP) . The SATP conditions are
298.15 K (250C) and 1 bar (105Pa) pressure the molar volume of ideal gas at SATP
conditions is 24.789 L mol-1.
Question 1: A manometer is connected to a gas containing bulb. The open arm reads 43.7 cm
where as the arm connected to the bulb read 15.6 cm. If barometric pressure is 743 mm Hg what
is the pressure of gas in bar?
Question 2: Side arm of an open-end mercury manometer is attached to a system. The level of
mercury in the arm attached to a system is 125 mm lower than the open end arm. Is thepressure higher than the atmospheric pressure or lower? What is the pressure of the system
if atmospheric pressure is 760 mm Hg?
Question 3: Open end mercury manometer is connected to a gas chamber containing a gas at a
pressure of 1.38 bar. The barometric pressure is 750 mm Hg. The arm connected to gas chamber
stands at 11.3 cm. What is the reading in open arm of the manometer?
Question 4: A student reported the gas pressure as 1.4 105kgm-1s-2. Report this value in kPa,
atm, mm of Hg and bar.
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GAS LAWS
BOYLES LAWThis law describes the pressure-volume relationship of gases at constant temperature. It was given
by Robert Boyle.The volume of a fixed mass of a gas is inversely proportional to its pressure at constant
temperature.
Mathematically, the law may be expressed as: (Temperature and mass constant)
Or where k1 is constant of proportionality Or PV=k1=constant.
Or P1V1=P2V2
For a given amount of the gas, the product of pressure and volume is constant at constant
temperature.
GRAPHICAL REPRESENTION OF BOYLES LAW
Q. No. 1 A balloon is filled with hydrogen at room temperature. It will burst if pressure exceeds
0.2 bar. If at 1 bar pressure the gas occupies 0.27 L Volume; upto what volume can the
balloon be expanded?
2. In a J tube partially filled with mercury the volume of air column is 4.2 mL and the mercurylevel in the two limbs is same. Some mercury is, now added to the tube so that the volume of
air enclosed in shorter limb is now 2.8 mL. What is the difference in the levels of mercury in
this situation? Atmospheric pressure is reported to be 1.0 bar.
3. At constant temperature, a gas occupies a volume of 200 mL at a pressure of 0.720 bar. It issubjected to an external pressure of 0.900 bar. What is the resulting volume?
4. A vessel of 120 mL capacity contains a certain amount of a gas at 1.2 bar pressure and 35 0C.The gas is transferred to another vessel of volume 180 mL at 35
0
C. What would be itspressure?
CHARLES LAWThis law describes the volume-temperature relationship of gases at constant pressure. It was put
forwarded by Jacques Charles.
The volume of a fixed mass of a gas increases or decreases by 1/273.15 of its volume at 00C
for each degree rise or fall of temperature, provided pressure is kept constant.If V0 is the volume of given mass of the gas at 0
0C
Then, V1, the volume of the gas at 10 C = V0 +
15.273
V0 = V0
15.273
11
P
IV
P
IkV 1
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V2, The volume of the gas at 20 C = V0
15.273
21
Vt, the volume of the gas at t0 C = V0
15.273
t1
ALTERNATIVE STATEMENT OF CHARLESS LAWWe have already derived the relationship between the volume of a given mass of the gas at t0 C
(Vt) and that at 00C (V0).
Where T is corresponding temperature in kelvin scale.
Or
Thus, V = k2T or V T (at constant P and n)
An alternative statement of Charles law as :The volume of the fixed mass of a gas at constant pressure is directly
proportional to the temperature in Kelvin scale.Let V1 be the volume of a certain mass of a gas at temperature T1 and at pressure P. If temperature
is changed to T2 keeping pressure constant, The volume changes to V2. The relationship between
four variables V1, T1, V2 and T2 is:
2
2
1
1
T
V
T
V (P and n are constant)
GRAPHICAL REPRESENTATION OF CHARLES LAWThe law can also be illustrated by volume-temperature curves. Plot of volume of a given of a gasagainst temperature at constant pressure. We get a graph as shown
GAYLUSSACS LAW OR AMONTONS LAWThis law describes the pressure-temperature relationship of gases at constant volume. It states that
pressure of a fixed mass of a gas at constant volume is directly proportional to the temperature in
Kelvin scale.
Mathematically, PT (At constant n, V)
QUESTIONS
15.273
t1VV 0t
0
100
15.273
15.273
T
TV
tV
0
0
t
1
T
V
T
V )nandPttanconsat(kttancons
T
V2
2
2
1
1
T
P
T
P
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1. A sample of a gas is found to occupy a volume of 900 cm3 at 270 celsius. Calculate thetemperature at which it will occupy a volume of 300 cm3, provided the pressure is kept
constant.
2. A steel tank contains air at a pressure of 15 bar at 200C. The tank is provided with a safetyvalue which can withstand a pressure of 35 bar. Calculate the temperature to which the tankcan be safety heated.
3. A sample of a gas occupies 10 L at 1270C and 1 bar pressure. The gas is cooled to 730C atthe same pressure. What would be the volume of the gas?
4. A vessel of capacity 400 cm3 contains hydrogen gas at 1 atm. Pressure at 70C. In order toexpel 28.57 cm3 of the gas at the same pressure to what temperature, the vessel should be
heated?
5. A flask having a volume of 250.0 mL and containing air is heated to 100 0C and sealed. Thenthe flask is cooled to 250C, immersed in water and opened. What volume of water will be
drawn back into the flask, assuming the pressure remaining constant?
AVOGADROS LAWEqual volumes of all the gases under similar conditions of temperature and pressure contain equal
number of molecules.
Ideal Gas Equation:- It is combination ofBoyles law, Charles law and Avogadros
law.
According to Boyles law V P
1(at constant T and n)
According to Charles law V T (at constant P and n)
According to Avogadros lawV = n (at constant T and P)
or PV nT ; PV nRT
Where, R is constant of proportionality and is known as universal gas constant. It is also known
as equation of state for ideal gas.
RELATIONSHIP BETWEEN DENSITY AND MOLAR MASS
,
Where d is the density of gas
V
w
Volume
Massd
R represents work done per degree per mole.
NUMERICAL VALUE OF R
(i) for 1 mol of gas
= 8.3144 Pa m3 K-1 mol-1 or8.3144 JK-1mol-1.
(ii) When pressure is expressed in atmosphere and volume in liters. =
0.0821LatmK-1
mol-1
.
(iii) When pressure is expressed in bar and volume in dm3
.
P
nT
V
RTM
wnRTPV
M
dRTPor
MV
wRTP
nT
PVR
)K15.273(x)mol1(
)m10x414.22()Pa101325(R
33
)K15.273(x)mol1(
)litres(414.22x)atm(1R
T
PVR
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We know that one mole of any gas at 1 bar pressure and 273.15 K (new S.T.P. conditions)
occupies a volume of 22.7 dm3 or 22.7 L. Then
= 0.0831 bar dm3
K-1
mol-1
.
= 1.99 calories K-1
mol-1 or = 2 cal K-1 mol-1.
Q.No.1: A sample of nitrogen gas occupies a volume of 320 cm3 at S.T.P. Calculate its
volume at 660C and 0.825 bar pressure.
2. 1.0 mol or pure dinitrogen gas at SATP conditions was put into a vessel of volume 0.025 m 3,maintained at the temperature of 50C. What is the pressure of the gas in the vessel?
3. How many moles of oxygen are present in 400 cm3 sample of the gas at a pressure of 760 mmof Hg at a temperature of 300 K. (The value of R is given to be 8.31 kPa dm3 k-1 mol-1).
4. A discharge tube containing nitrogen gas at 250C is evacuated till the pressure is 2x10-2 mm. Ifthe volume of discharge tube is 2 liter, calculate the number of nitrogen molecules still present
in the tube.
5. The density of certain gaseous oxide at 1.5 bar pressure at 100C is same as that dioxygen at200C and 4.5 bar pressure. Calculate the molar mass of gaseous oxide.
6. Density of a gas is found to be 5.46g/dm3 at 270C and 2 bar pressure. What will be its densityat S.T.P.?
7. Temperature and pressure in Chandigarh are respectively 350 C and 740 mm whereas atShimla these are 100 C and 710 mm. calculate the ratio of densities of air at Chandigarh (d1)
and at Shimla (d2).
8. 2.9 g of a gas at 950 C occupies same volume as 0.184 g of H2 at 170C at the same pressure.What is the molar mass of the gas?
9. Calculate the temperature of 4.0 mol of gas occupying 5 dm3 at 3.32 bar pressure. (R=0.083bar dm3 K-1 mol-1).
10.An air bubble has a radius of 0.50 cm at the bottom of the water tank where the temperature is280 K and the pressure is 280 kPa. When the bubble rises to the surface, the temperature
changes to 300 K and pressure is 100 kPa. Calculate the radius of the bubble at the surface.
11.A gas balloon having a volume of 10.0 L at 250C and 1 atm. Pressure rises to an altitudewhere the temperature is100C and pressure 500 mm of Hg. What is the volume of balloon at
this altitude?
12.The pressure of the atmosphere 100 miles above the earth is about 2x10-6 mm of Hg and thetemperature is180
0
C. Calculate number of gaseous molecules per mL at this altitude.
DALTONS LAW OF PARTIAL PRESSURESThe law was given by John Dalton
At constant temperature, the pressure exerted by a mixture of two or more non-reacting
gases enclosed in a definite volume, is equal to the sum of the individual pressures which
each gas would exert if present alone in the same volume at the same temperature.The individual pressures of gases are known as partial pressures. P=P1+P2+P3+ (T, V
are constant)
Question1:-
)K15.273(x)mol1(
)dm(414.22x)bar(1R
3
7
7
10x183.4
10x314.8R
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1. A certain quantity of a gas occupies a volume of 0.1 L when collected over water at 288 K anda pressure
0.92 bar. The same gas occupied a volume of 0.085 L at S.T.P. in dry conditions. Calculate the
aqueous tension at 288 K.
2. A sample of O2 is collected over water at 220C and 748 torr pressure. The volume of the gascollected is 82.0 cm3. How many grams of oxygen are present in the gas? Vapour pressure ofwater at 220C is 19.8 torr.
3. A 2 L flask containing nitrogen at 60 cm pressure is put into a communication with 4 L flaskcontaining carbon monoxide at 48 cm pressure. If the temperature is kept constant. Calculate
the final pressure of the mixture.
4. A flask of 1.5 L capacity contains 400 mg of O2 and 60 mg of H2at 1000C. Calculate the totalpressure of the gaseous mixture. . If the mixture is permitted to react to form water vapours
at 1000C, what materials will be left and what will be their partial pressures.
5. A gaseous mixture contains 5.6 g of carbon (II) oxide and rest carbon (IV) oxide when it isenclosed in a vessel of 10 d m3 at 298 K it recorded a pressure of 2.0 bar . What is the
partial pressure of each oxide of carbon?6. Two bulbs A and B of equal volumes connected through stopcock (kept open) contained 0.7
mol of H2 gas at pressure of 0.5 atm and 270C. Calculate final pressure and molar
distribution of H2 in two bulbs.
7. Pressure of 1 g of ideal gas A at 270C is found to be 2 bar. When 2g of another gas isintroduced in the same flask at same temperature the pressure becomes 3 bar . Find the
relationship between their molar masses.
KINETIC MOLECULAR THEORY OF GASESThe various postulates of the theory are:
1. All gases are made up of very large number of extremely small particles called molecules.2. The molecules are separated from one another by large spaces so that the actual volumeoccupied by the molecules is negligible as compared to the total volume of the gas.
3. The molecules are not at rest but posses rapid random motion. During their motion, theycollide with one another and also against the walls of the container.
4. The pressure of the gas is due to bombardment of the gas molecules against the walls of theconainer.
5. The collisions of the molecules with each other and with the walls of the container areperfectly elastic, i.e. there may be redistribution of energy during such collision.
6. There are no attractive or repulsive forces between the molecules of the gas. They arecompletely independent of each other.
7. At any instant, different molecules possess different velocities and hence, different energies.However, the average kinetic energy of the molecules is directly proportional to the absolutetemperature.
(i) Most probable speed ( or ump). It is the speed possessed by maximum fraction of molecules
of the gas at particular temperature. It is given by the expression
(or ump)= (M is molar mass of gas)
(ii) Average speed ( or uav). It is the average of the speeds of different molecules of the gas
at particular temperature. If v1, v2, v3,.are speeds of different molecules, then
M
RT2
v
N....vvv)uor(v 321av
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Average speed is given by the expression
(iii)Root mean square speed (ur.m.s.). It is the square root of the mean of the squares ofspeeds of various molecules of the gas at a given temperature. If v1, v2,v3.. are speeds
of differ
ent molecules, then
Root mean square speed is given by the expressions
or
Relationship between different types of speeds.The different types of speeds are related as :
KINETIC GAS EQUATION AND AVERAGE KINETIC ENERGY OF GAS
MOLECULESPV=1/3 MnU
2.
The average kinetic energy of gas molecules can be calculated on the basis of kinetic gas
equation. P=1/3 (N/V)mu2
For one mol of the gas, PV=1/3NAmu
2[N=NA for 1 mol]
Also for 1 mol of ideal gas, PV=RT
1/3NAmu2=RT
Now mu2 represents average kinetic energy (K. E.) of the molecules.
Q.No. 1 Calculate (i) average kinetic energy of 32 g of methane molecules at 270C R=8.314
JK-1 mol-1.
(ii) Root mean square speed and (iii) Most proable speed of methane molecules at 270C.
(2) Calculate the root mean square speed of oxygen molecule at 270
C.(3) At what temperature rms velocity of SO2 is equal to that CH4 at 27
0C?
Derivation of Gas Laws from Kinetic Gas equation
Boyle,s Law and Charles Law
At constant Temp.
M
RT8)uor(v av
N
.....vvvu
23
22
21
.s.m.r
M
PV3or
M
RT3ur .a.m.r
.224.1:128.1:1::u:v: .a.m.r
RT2
3mu
2
1orRTmuN
3
2X
2
1or 22A
kT2
3.E.KorRT
2
3.E.K
2Num3
1PV 2mu
2
1X
3
2PV KEX
3
2 TKE
T3
2PV T
3
2XKPVor
P
T
3
2XKV
contsantTXXK 3
2
PXtconsV
1tan
D
P3
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Or PV = constant (Boyles Law)
At constant pressure
V= constant T, Or V/T= constant
(Charles Law)
ttancons
P3
2XK
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GASEOUS STATEII
Dalton Law of Partial pressureLet there be two gases in a container having root mean square velocity u1 and u2 number of
molecules N1 and N2 and mass of molecules m1 and m2.
Pressure due to molecules of gas A Pressure due to Gas B
Molecules of two gases do not interfere each other in their motion so total pre ssure
=PA+PB (Daltons Law of partial pressure)
MAXWELLS DISTRIBUTION OF VELOCITIESMolecules of a gas are in random motion and their velocities go on changing due to molecular
collision so it is not possible to find out actual velocity of a molecule but it is possible to find outfraction of total number of molecules having a certain velociy.
Maxwell plotted vs velocity
Maxwell distribution of velocitiesImportant features of Maxwell distribution curve are
(i) The fraction of molecules with very low or very high velocities (speeds) is very small.(ii)The fraction of molecules possessing higher and higher speeds goes on increasing till it
reaches the peak and thereafter it starts decreasing.
(iii)The maximum fraction of molecules possesses a velocity (or speed) corresponding to the peak
of the curve. This speed corresponding to the peak in the curve is referred to as mostprobable speed.
(iv)The area under the curve gives the total number of gas molecules.IDEAL GASES AND REAL GASESIdeal gases are those which obey gas laws under all conditions of temperature and pressure.
No gas is an ideal gas.
REAL GASES:- All gases obey gas Laws at high temperature and low pressure but do not obey
gas laws at low temperature and high pressure . Such gases are known as real gases.
Compressibility factor Z
(i) For an ideal gas, as PV=nRT, Z=1
V
uNm
3
1pA
12
11 V
uNm
3
1pB
12
22
V
uNm
3
1
v
uNmP
22
2212
1131
N
N
nRT
PVZ
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(ii)For a real gas, as PV nRT, Z 1. Hence, two cases arise:(a)When Z1, the gas is said to show positive deviation. This implies that the gas is less
compressible than expected from ideal behavior. The temperature at which a real gas behaveslike an ideal gas over an appreciable pressure range is called Boyle temperature or Boyle
point.
Cause of Deviation from ideal Behaviour.The derivation of the gas law (and hence of the ideal gas equation) is based upon the Kinetic
Theory of Gases which in turn is based upon certain assumptions. Thus, there must be something
wrong with certain assumptions. A careful study shows that at high pressure or low temperature,
he following two assumptions of Kinetic Theory of Gases is faulty:
(i) The volume occupied by the gas molecules is negligible as compared to the total volume ofthe gas.
(ii)The force of attraction or repulsion between the gas molecules is negligible.1
Equation of State for the Real gases (Vandar Waals equation). To explain the behavior of real
gases, J.D. Vandar Waal, in 1873, modified the ideal gas equation by applying appropriate
corrections so as to take into account
(i) the volume of the gas molecules and(ii)the forces of attraction between the gas molecules.He put forward the modified equation, known after him as van Waals equation. This equation for
1 mole of the gas is
And for n moles of the gas, it is
Where aand b are constants, called van der Waals constants. Their values depend upon the
nature of the gas.
Significance of Vander Waals constants(i) Vander Waals constant a. Its value is a measure of the magnitude of the attractive forces
between the molecules of the gas. Greater the value of a stronger are the intermolecular forces of
attraction.
(ii)Vander Waals constant b. Its value is a measure of the effective size of the gas molecules.Its value is equal to four times the actual volume of the gas molecules. It is called excluded
volume or co-volume.Units of Vander Waals constants(i) Units of a. As
=atm L2 mol-2 or bar dm6 mol-2
(ii)Units of b. As volume correction V = nb,Explanation of the behaviour of real gases by Vander Waals equation
RTbVV
aP
2
nRTnbV
V
anP
2
2
2
2
2
2
n
pXV
,V
anp
131 moldmormolLn
vb
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(i) At very low pressures, V is very large. Hence, the correction term a/V2 is so small that it canbe neglected. in comparison to V. Thus, Vander Waals equation reduces to the form PV=RT. i.e.
behave like ideal gases.
(ii)At moderate pressures, V decreases. Hence, a/V2 increases and cannot be neglected.However, V is still large enough in comparison to b so that b can be neglected. Thus, VanderWaals equation becomes
(iii)At high pressure, V is also small that b cannot be neglected in comparison to V. The factor
a/V2 is no doubt large but as P is very high, a/v 2 can be neglected in comparison to P. Thus,
Vander Waals equation reduces to the form
P(V-b)=RT Or PV=RT+pb
(iv)At high temperature, V is very high (at given pressure) so that both the correction factors (a/v 2and b) become negligible as in case (I) . Hence, at high temperature, real gases behave like ideal
gas.
Explanation of the exceptional behaviour of hydrogen and helium. For H2 and him, the
compressibility factor Z is always greater than 1 and increases with increase of pressure. This is
because H2 and He being very small molecules, the intermolecular forces of attraction in them are
negligible i.e. a is very very small so that a/V2 is negligible.
The Vander Waals equation, therefore, becomes
P (V-b)=RT Or PV=RT+ Pb
Thus PV/RT i.e. Z>1 and increases with increase in the value of P at constant T.
Q.No.1 Calculate the pressure exerted by 11.0 g of carbon dioxide in a vessel of 2 L capacity
at 370C, Given the Vander Waals constants are a=3.59 L2 atm mol-2 and b=0.0427 L
mol-1 . Compare the value with the calculated value if the gas were considered as
ideal.
Q.No.2. Calculate the temperature of 2 moles of sulphur dioxide gas contained in a 5 L vessel
at 10 bar pressure. Given that for SO2 gas, Vander Waals constants are :a=6.7 bar L2mol-
2 and b=0.0564 L mol-1.
Liquefaction of Gases and Critical TemperatureThe liquefaction of a gas takes place when the intermolecular forces of attraction become so high
that they blind the gas molecules together to form the liquid state. The intermolecular forces of
attraction can be increased either by increasing 2 the pressure so that the molecules come close
together or by cooling the gas so that the kinetic energy of the molecules decreases and they
become slower. Hence, a gas can be liquefied by cooling or by application of pressure or the
combined effect of both. The effect of temperature on the liquefaction of gases is found to be very
important as higher the temperature of the gas, more difficult it is to liquefy it and higher is the
pressure required.
RTV
PVorRTVV
aP
2
V
aRTPVor
RTV
a1
RT
PVor
RTV
a1Zor
RT
pb
1RT
PV
or RT
pb
1Zor
RT
Pb
1RT
PV
Or
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Critical temperature of a gas may be defined as that temperature above which it cannot be
liquefied howsoever high pressure may be applied on the gas.
Andrews Experiments on Critical phenomena. It is interesting to mention here that Andrews
in 1861 was the first to study the critical phenomena experimentally using CO 2 gas. He studiedthe effect of pressure on the volume of CO2 at different constant temperature. Some of the
isotherms thus obtained is
At the lowest temperature employed, i.e. 13.10C, at low pressure, CO2 exists as a gas, as shown at
the point A. As the pressure is increased, the volume of the gas decreases along the curve AB. At
the point B, liquefaction of the gas starts. Hence, volume decreases rapidly along BC because
liquid has much less volume than the gas. At the point C, liquefaction is complete. Now, the
increase in pressure has very little effect upon volume because liquids are very little compressible.
Hence, a steep curve CD is obtained. As the temperature is increased, horizontal portion becomes
smaller and smaller and 30.980C, it reduced at a point, P. This means that above 30. 980C, the gas
cannot be liquefied at all, however high pressure may be applied. Thus, 30.980C is the critical
temperature. This corresponding pressure to liquefy the gas at the critical temperature is its critical
pressure, Pc (i.e. 73.9 atm). The volume occupied by 1 mole of the gas under these conditions is
its critical volume, Vc(i.e., 95.6 ml).
Joining the end point of the horizontal portion of the different curves, we get a dome-shaped curve
as shown in Figure 5.31. We observe that a point like A in the Fig. Represents gaseous state. A
point like D represents a liquid state and a point within the dome shaped area represent existence
of liquid and gaseous CO2 in equilibrium. All the gases on isothermal compression show the samebehaviour as shown by CO2.
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MCQ with only one Answer
1. At constant temperature, the productof pressure and volume of a givenamount of a gas is constant. This is
(a) Gay-Lussac Law(b) Charles Law(c) Boyles Law
(d) Pressure Law
2. Charles law is representedmathematically as:
(a) V1=KV0t
(b)
(c) (d)
3. Correct gas equation is:(a) (b)
(c)
(d)
4. In general gas equation, PV=nRT, V isthe volume of:
(a) one moles of a gas
(b) any amount of a gas(c) one moles of a gas
(d) one gram of gas
5. In the equation of state of an ideal gasPV=nRT, the value of universal gas
constant would depend only on:
(a) The nature of the gas(b) The units of measurement
(c) The pressure of the gas
(d) The temperature of the gas6. The value of the gas constant per
degree per mol is approximately:
(a) 1 cal (b) 2 cal
(c) 3 cal (d) 4 cal7. 10 g of a gas at atmospheric pressure
is cooled from 2730C to 0
0C keeping
the volume constant; its pressurewould become:
(a) atm (b) 1/273 atm
(c) 2 atm (d) 273 atm
8. 300 mL of a gas at 270C is cooled to 30C aconstant pressure: the final volume is:
(a) 540 mL (b) 135 mL(c) 270 mL (d) 350 mL9. 273 mL of a gas at S.T.P. was taken to 270C
and 600 mm pressure. The final volume of the
gas would be:(a) 273 mL (b) 300 mL
(c) 380 mL (d) 586 mL
10.The density of a gas is 1.964 g dm-3 at 273 Kand 76 cm Hg. The gas is:(a) CH4 (b) C2H6
(c) CO2 (d) Xe
11.The density of the gas is equal to:(a) nP (b) MP/RT
(c) P/RT (d) M/V
12.A bottle of dry ammonia and a bottle of dryhydrogen chloride connected through a longtube are open simultaneously at both the ends
The white ring first formed will be:
(a) At the centre of the tube(b) Near the ammonia bottle
(c) Near the HCl bottle
(d) Throughout the length of the tube
13.Which of the following gases will have thehighest rate of diffusion?
(a) O2 (b) CO2
(c) NH3 (d) N214.Equal masses of methane and oxygen are
mixed in an empty container at 250C. The
fraction of the total pressure exerted by oxygenis:
(a) 1/3 (b) 1/2
(c) 2/3 (d)15.50 mL of a gas A diffusion through a
membrane in the same time as for the diffusion
of 40 mL of gas B under identical conditions of
pressure and temperature. If the molecular massof A is 64, that of B would be:
(a) 100 (b) 250 (c) 200 (d) 80
16.3.2 g of oxygen (at. Mass =16) and 0.2 g ofhydrogen (at. Mass =1) are placed in a 1.12
t
KVV 0t
t
2731VV 0t
273
t1VV 0t
2
1
22
11
TT
VPVP
2
12
1
21PTV
PTV
2
22
1
11
V
TP
V
TP
2121
21 PPTT
VV
298
273X
3
1
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litre flask at 00C. The total pressure of
the gas mixture will be
(a) 1 atm (b) 2 atm(c) 3 atm (d) 4 atm
17.The density of a gas at 27
0
C and 1 atmis d. Pressure remaining constant, atwhich of the following temperature
will its density become 0.75d?
(a) 200C (b) 30
0C
(c) 400K (d) 300K
18.Equal masses of ethane and hydrogenare mixed in an empty container at
250C . The fraction of the total
pressure exerted by hydrogen is:
(a) 1:2 (b) 1:1
(c) 1:16 (d) 15:1619.The kinetic theory of gases predicts
that total kinetic energy of a gas
depends on:
(a) Pressure of the gas(b) Temperature of the gas
(c) Volume of the gas
(d) Pressure, temperature and volumeof the gas
20.Deviations from ideal behaviour willbe more if the gas is subjected to:
(a) Low temperature and low pressure
(b) high temperature and low pressure
(c) Low temperature(d) High temperature
21.The RMS speed at NTP of the speciescan be calculated from the expression:
(a) (b)
(c) (d) All
22.Non-ideal gases approach idealbehaviour:
(a) High temperature and highpressure
(b) High temperature and low pressure
(c) Low temperature and low pressure
(d) Low temperature and low pressure
23.The critical temperature of gas is thattemperature:(a) Above which it can no longer remain in the
gaseous state(b) Above which it cannot be liquefied bypressure
(c) At which it solidifies
(d) At which the volume of the gas becomezero
24.The units of a in van der Waals equation are
(a) atom litre2
mol-2
(b) atom litre mol-2
(c) atom litre mol-1
(d) atom litre2 mol-1
25.The value of van der Waals constant for gasesO2. N2 NH3 and CH4 are 1.360, 1.390, 4.170
and 2.253 litre2
are mol-2
respectively. The gaswhich can be most easily liquefied is:
(a) O2 (b) N2
(c) NH3 (d) CH426.According the kinetic theory of gases for a di-
atomic molecules:
(a) the pressure exerted by the gas is
proportional to the mean square speed by themolecules
(b) the pressure exerted by the gas is
proportional to the root mean square speed ofthe molecules
(c) the root mean square speed is inversely
proportional to the temperature(d) The mean translational K.E. of the
molecules is directly proportional to the
absolute temperature.
27.The compressibility factor of a gas is defined asnRT
PV
Z The compressibility factor of an ideal gas is :
(a) 0 (b) 1 (c) -1 (d) infinity
28.An ideal gas is one which obeys:(a) gas laws (b) Boyles law(c) Charles law (d) Avogadros law
M
PV3
d
P3
M
RT3
nRTnbVV
anP
2
2
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29.16 g of oxygen and 3 g of hydrogenare mixed and kept at 760 mm
pressure at 00C. The total volume
occupied by the mixture will be
nearly:(a) 22.4 litre (b) 33.6 litre(c) 44800 mL (d) 4480 mL
30.Which of the following mixture ofgases at room temperature does notfollow Daltons law partial pressure?
(a) NO2 and O2 (b) NH3 and
HCl
(c) CO and CO2 (d) SO2 andO2
31.In which of the following pairs thegaseous species diffusion through aporous plug with the same rate of
diffusion?
(a) NO, CO (b) NO, CO2
(c) NH3,PH3 (d) NO,C2H6
32.At S.T.P., the order of mean squarevelocity of molecules of H2, N2, O2and HBr is:
(a) H2>N2>O2>HBr (b)
HBr>O2>N2>H2
(c) HBr>H2>O2>N2 (d)N2>O2>H2>HBr
33.A balloon filled with N2O is prickledwith sharper point and plunged into atank of CO2 under the same pressure
and temperature. The balloon will:
(a) be enlarged(b) shrink
(c) collapse completely
(d) remain unchanged in size34.If the universal gas constant is 8.3
joule mol-1
K-1
and the Avogadrosnumber 6X1023. The mean kinetic
energy of the oxygen molecules at
327o
C will be:(a) 415X10
-23joule (b)
2490X10-22
joule
(c) 1245X10-23
joule (d) 830X10-
22joule
35.If increase in temperature and volume of anideal gas is two times, then initial pressure of P
changes to;(a) 4P (b) 2P (c) P (d) 3P
36.At what temperature would the RMS speed of agas molecule have twice its value at 100
oC?
(a) 4192 K (b) 1492 K
(c) 9142 K (d) 2491 K
37.At a temperature T K, the pressure of 4 g argonin a bulb is P. The bulb is put in a bath having a
temperature higher by 50 K than the first one
0.8 g of argon has had to be removed to
maintain original pressure. The temperature Tis equal to:
(a) 510 K (b) 200 K
(c) 100 K (d) 73 K38.At 250C and 730 mm pressure, 380 mL of dry
oxygen was collected. If the temperature is
constant, what volume will the oxygen occupy
at 760 mm pressure?(a) 265 mL (b) 365 mL
(c) 569 mL (d) 621 mL
39.4The root mean square velocity of an ideal gasat constant pressure varies with density as:
(a) d2
(b) d (c) (d)
(d) Cannot be predicated unless the volumes of
the gases are given40.The coordination number of a metal crytallising
to a hexagonal close packing (hcp) structure is
:-(a) 12 (b) 4
(c) 8 (d) 6
41.The compressibility of a gas is less than unityat S.T.P. Therefore :
(a) Vm>22.4 litre (b) VmT(N2)(c) T(H2)
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(a) Zero (b) 3408 J
(c) 2 cal (d) 5.66 10-21
J
44.Densities of two gases are in the ratio1:2 and their temperature are in the
ratio 2:1; then the ratio of theirrespective pressures is:(a) 1:1 (b) 1:2 (c) 2:1 (d) 4:1
45.4.4 g of a gas at STP occupies avolume of 2.24 L. The gas can be:(a) O2 (b) CO
(c) NO2 (d) CO246.At 00C and one atm. Pressure a gas
occupies 100 cc. If the pressure isincreased to one and a half time and
temperature is increased by one third
of absolute temperature, then finalvolume of the gas will be:
(a) 80 cc (b) 88.9 cc
(c) 66.7 cc (d) 100 cc
47.Pressure of a mixture of 4 g of O2 and2 g of H2 confined in a bulb of 1 litre
at 00Cis:
(a) 25.215 atm (b) 31.205atm
(c) 45.215 atm (d) 15.210
atm
48.Density ratio of O2 and H2 is 16:1. Theratio of their rms velocity will be:
(a) 4:1 (b) 1:16
(c) 1:4 (d) 16:149.The rate of diffusion of a gas having
molecular weight just double of
nitrogen gas is 56 mL s-1
. The rate ofdiffusion of nitrogen will be:
(a) 79.19 mL s-1
(b) 112.0 mL s-1
(c) 56 mL s-1
(d) 90.0 mLs
-1
50.The density of air is 0.00130 g/mL.The vapour density of air will be:
(a) 0.00065 (b) 0.65(c) 14.4816 (d)
14.56
51.If 300 mL of gas at 270C is cooled to 70C atconstant pressure, its final volume will be:
(a) 135 mL (b) 540 mL(c) 350 mL (d) 280 mL
52.The van der Waals equation reduces itself tothe ideal gas equation at :(a) high pressure and low temperature
(b) low pressure and low temperature
(c) low pressure and high temperature(d) high pressure alone
53.The maximum number of molecules is presentin:
(a) 15 L of H2gas at STP(b) 5 L of N2 gas at STP
(c) 0.5 g of H2gas
(d) 10 g of O2
gas54.The root mean square velocity of one mole of a
monoatomic gas having molar mass M is UrmsThe relation between average kinetic energy
(E) of the gas and Urms is :
(a) (b)
(c) (d)
55.A certain sample of a gas has a volume of 0.2litre measured at 1 atm pressure and 00C . At
the same pressure but 2730C , its volume wil
be:(a) 0.4 L (b) 0.8 L
(c) 27.8 L (d) 55.6 L
56.The pressure exerted by 1 mole of methane in a0.25 litre container at 300 K using van der
Waals equation (given a=2.253) atm L2
mol-2
b=0.0428 L mol-1
) is :
(a) 82.82 atm (b) 152.51 atm(c) 190.52atm (d) 70.52 atm
57.A gas can be liquefied:(a) above its critical temperature
(b) at its critical temperature(c) below its critical temperature
(d) at any temperature
M2
E3U rms
M3
E2U rms
M
E2U rms M3
EU rms
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58.If the absolute temperature of a gas isdoubled and the pressure is reduced to
one half, the volume of the gas will:(a) remain unchanged
(b) be doubled(c) increase fourfold(d) be halved
59.To what temperature must a neon gassample be heated to doubled thepressure, if the initial volume of a gas
at 750C is decreased by 15%?
(a) 3190C (b) 592
0C
(c) 1280C (d) 60
0C
60.In Which of the following crystalsalternate tetrahedral voids are
occupied(a)NaCl (b) ZnS
(c) CaF2 (d) Na2O
ANSWERS
1 c 16 d 31 b 46 b
2 d 17 c 32 d 47 a
3 a 18 d 33 a 48 c
4 a 19 b 34 d 49 a
5 b 20 a 35 c 50 d
6 b 21 d 36 c 51 d
7 a 22 b 37 d 52 c8 c 23 b 38 b 53 a
9 c 24 d 39 d 54 b
10 b 25 a 40 a 55 a
11 c 26 c 41 b 56 a
12 c 27 d 42 c 57 c
13 c 28 b 43 d 58 c
14 a 29 a 44 a 59 a
15 a 30 c 45 d 60 b