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SUMMARY
*
1.
Gas Laws:
Boyle's law: For a given mass of an ideal gas at constant temperature, the volume of a gas is
inversely proportional to its pressure, i.e.
Vα
1
P
or PV = constant  P1V1 = P2V2
P1 P2
m
P
(i) PV  P  = constant 
= constant or   

1
2

m
Where density,
  , and m = constant
V
(ii) As number of molecules per unit volume
n 
N
V
 V 
 N
(iii) PV  P  =
n
constant 
N
also N = const.
n
P1
P2
P
= constant or n  n
n
1
2
(iv) Graphical representation: (If m and T are constant)
2.
Charle's law: At constant pressure, the volume of the given mass of a gas is directly proportional
to its absolute temperature.
i.e. V  T 
(i)
V1
V2
V
= constant = T  T
1
2
T
V
m

m 

= constant  V 

T T
ρ 

or T = constant = 1T1  2T2
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(ii)
3.
Graphical representation: (If m and P are constant)
Gay-Lussac`s Law or pressure law:
According to it for a given mass of an ideal gas at constant volume, pressure of a gas is
directly proportional to its absolute temperature.
i.e.
P  T or
(i)
*
P
= constant
T

P1 P2

T1 T2
Graphical representation: (If m and V are constant)
Avogadro's law: Equal Volume of all the gases under similar conditions of temperature and
pressure contain equal number of molecules.
i.e. N1 = N2
Avogadro Number: The number of particles (atoms or molecules) in one mole of substance
(gas) is called Avogadro number (NA) which has a magnitude
NA = 6.023  1023 mol-1
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*
Equation of State OR Ideal Gas Equation
The equation which relates the pressure (P), volame (V) and temperature (T) of the given
state of an of an ideal gas is known as ideal gas equation or equation of state.
Ideal gas equation is PV = µRT
*
where µ =
R=
=
Different forms
number of mole
universal gas constant
8.314 J.mol-1 K-1
of Ideal gas-state-equation
 N 


R
PV  µRT  
 kB 
 RT  Nk BT 
 NA 
 NA

(i)
where kB = Boltzmann`s constant
= 1.38  10-23 JK-1
(ii)
P
N
k BT  nk BT
V
where n 
(iii)
PV 
N
V
=
number density of gas
=
number of molecules per unit volume

M
M 
RT  µ 

Mo
Mo 

where Mo =
P
*
molar mass of the gas
M RT ρRT
m
= ( ρ 
density of the gas)

V
V Mo Mo
The work done during the change in volume of the gas:
Vf
It can be obtained from the graph of P – V W   P dν
Vi
*
Assumption of Ideal gases (or kinetic theory of gases)
Kinetic theory of gases relates the macroscopic properties of gases (such as pressure, temperature
etc.) to the microscopic properties of the gas molecules (such as momentum, speed, kinetic
enengy of molecule etc.)
Assumptions:
1.
Gas is made up of tiny particles, These particles are called molecules of the gas.
2.
The molecules of a gas are identical, spherical, rigid and perfectly elastic point masses.
3.
The molecules of a gas perform incessant random motion.
4.
The molecules of a gas follow Newton's laws of motion.
5.
The number of molecules in a gas is very large.
6.
The volume of molecules is negligible in comparision with the volume of gas.
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7.
8.
*
Inter molecular forces act only when two molecules come close to each other or collide.
The time spent in a collision between two moleules is negligible in comparision to time
between two successive collisions. The collisions between the molecules and between a
molecule and the wall of a container are elastic.
Pressure of an Ideal gas
P
*
1
1
ρ  ν2  
ρ ν 2 rms
3
3
where  = density of the gas
v2rms = <2> = mean squre velocity of molecule
Relation between pressure and kinetic energy
P
1
ρ ν 2 rms        (i)
3
K.E. per unit volume is
E =
1 M 2
1
 rms =
 2rms        (ii)
2 V
2
From equation (i) and (ii)
P
*
2
E
3
Root mean square speed (vrms) :
It is defined as the square root of mean of squares of the speed of the speed of different
molecules.
i.e. rms =
ν12  ν22  ν32  ........  νN2
N
From the expression of pressure
1
 2rms  rms =
3
P =
3RT
=
M
=
where ρ 
m
3P
=
ρ
3PV
M
3k BT
m
M
= density of the gas
V
M
 mass of each molecule
NA
-
νrms  T
-
with increase in molecular weight rms speed of gas molecule decreases as vrms 
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1
M
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*
rms speed of gas molecules does not depend on the pressure of gas (If temperature remains
constant)
At T= 0 K , vrms= 0, i.e. the rms speed of molecules of a gas is zero at O K. This temperature
is called absolute zero.
Kinetic interpretation of temperature
Kinetic energy of of 1 mole ideal gas
E
1
1  3RT  3
M ν 2 rms  M 
  RT
2
2  M  2
-
For 1 molecule E 
3
kBT , kB = Boltzmann`s constant
2
-
For N molecule E 
3
N kBT
2
-
Kinetic energy per nolecule of gas does not depend upon the mass of the molecule but only
depends on the temperature.
Mean free path:The distance travelled by a gas molecule between two successive collisions is known as free path.
The average of such free paths travelled by a molecule is called mean free path.
mean free path,
*
Total distance travelled by a gas molecule between two successive collisions
 = Total number of collisions
1
2 nd 2
-
 
-

-
Collision frequency

1
k BT
P 

 P  nk BT  n 

2
2 
k BT 
2 nπd
2 π Pd 
= number of collisions per second.

νrms


*
-
Degrees of Freedom :
The term degrees of freedom of a molecule or gas are the number of independent motions that
a molecule or gas can have.
The independent motion of a system can be translational, rotational or vibrational or any combination
of these.
Degress of freedom,
f = 3A - B; where A = Number of independent particles,
B = Number of independent restrictions
monoatomic gas  3 degrees of freedom
( All translational)
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-
Diatomic gas
-
triatomic gas
 5 degrees of freedom
( 3 translational + 2 rotational)

(Non-linear)
6 degrees of freedom (at room temperature)
( 3 translational + 3 rotational)

8 degrees of freedom (at very high temperature)
(3 translational + 3 rotational + 2 vibrational)
*
Law of equipartition of energy (Boltzmann law)
According to this law, for any system in equilibrium, the total energy is equally distributed
among its various degrees of freedom and each degrees of freedom is associated with energy
-
1
k B T where kB = Boltzmann`s constant.
2
At a given temperataue T, all ideal gas molecules no matter what their mass have the same average
3
k BT
2
At same temperature gases with different degrees of freedom (i.e. H2 and He) will have different
translational kinetic energy =
-
average energy 
f
k BT
2
( f = degress of freedom different for different gases.)
f
k BT
2
-
The total energy associated with each modlec 
*
Specific heat of a gas :
-
Specific heat at constant volume (Cv)
-
The amount of heat required to change the temperature of l mole of gas by 1 K, keeping its
volume constant, is called specific heat of the gas at constant volume.
-
Specific heat at constant volume (Cp)
The amount of heat required to change the temperature of l mole of gas by 1 K, keeping its
pressure constant, is called specific heat of the gas at constant pressure.
Molar specific heat:
-
The quantity of heat required to change the temperature of 1 mole of gas by 1 K (or 1o C) is
called molar specific heat of the gas.
Ratio of CP and CV is .
f

 1 R

C
2
2 
 P 
 1
f
CV
f
R
2

1
 f
 
 CV  f R , C P    1 R 
2
2
 

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MCQ
Choose the correct alternative from given options.
1.
Volume, pressure and temperature of an ideal gas are V, P and T respectively. If mass of molecule
is m, then its density is [ kB = Boltzmann`s constant]
P
(A) k T
B
2.
3.
4.
5.
6.
Pm
(B) k T
B
(B)
9.
10.
p
8
(C) 2 P
(D) P
The equation of state for 5 g of oxygen at a pressure P and temperature T, when occupying a
volume V, will be
 5 
(A) PV    RT
 32 
8.
(D) mk B T
The temperature of an ideal gas at atmospheric pressure is 300 K and volume 1 m3. If temperature
and volume become double, then pressure will be
(A) 4  105 Nm–2
(B) 2  105 Nm–2
(C) 1  105 Nm–2
(D) 0.5  105 Nm–2
At 100 K and 0.1 atmospheric pressure, the volume of helium gas is 10 litres. If volume and
pressure are doubled, its temperature will change to
(A) 127 K
(B) 400 K
(C) 25 K
(D) 200 K
What is the mass of 2 litres of nitrogen at 22.4 atmospheric pressure and 273 K.
(R = 8.314 Jmol k-1)
(A) 14  22.4 g
(B) 56 g
(C) 28 g
(D) None of these.
An electron tube was sealed off during manufacture at a pressure of 1.2  10-7 mm of mercury at
270C. Its volume is 100 cm3. The number of molecules that remain in the tube is _______
(density of mercury is 13.6 gcm–3)
(A) 3.9  1011
(B) 3  1016
(C) 2  1014
(D) 7  1011
A vessel contains 1 mole of O2 gas (relative molar mass 32) at a temperature T. The pressure of
the gas is P. An identical vessel containing 1 mole of He gas (relative molar mass 4) at a
temperature 2T has pressure of ..........
(A) 8 P
7.
.P
(C) k TV
B
5
(B) PV    RT
 16 
5
(C) PV    RT
2
(D) PV = 5 RT
A gas at 1 atmosphere and having volume 100 ml is mixed with another gas of equal moles at
0.5 atm and having volume 50 ml in flask of one litre, what is the final pressure?
(A) 0.125 atm
(B) 0.75 atm
(C) 1 atm
(D) 0.5 atm
PV
The quantity k T represents
B
(A) mass of gas
(B) number of moles of gas
(C) number of molecules in gas
(D) K. E. of gas
Equation of gas in terms of pressure (P), absolute temperature (T) and density () is
P1T1 P2T2
(A) ρ  ρ
1
2
P1
P2
(B) T ρ  T ρ
1 1
2 2
(C)
P1ρ 2 P2ρ1

T1
T2
P1ρ1 P2ρ 2
(D) T  T
1
2
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11.
02 gas is filled in a vessel. If pressure is doubled, temperature becomes four times, how many times
its density will become.
(A) 4
12.
13.
(B)
1
4
15.
1
2
(A) Varies inversely as the square of its mass
(B) Varies inversely as its mass
(C) is independent of its mass
(D) Varies linearly as its mass
If pressure of a gas contained in a closed vessel is increased by 0.4% when heated by 10 C the
initial temperature must be.
(B) 250 C
(C) 250 K
(D) 2500 K
To decrease the volume of a gas by 5% at constant temperature the pressure should be
(A) Incseased by 5.26%
(B) Decreased by 5.26%
(C) Decreased by 11%
(D) Increased by 11%
A gas at the temperature 250 K is contained in a closed vessel. If the gas is heated through
1 K, then the percentage increase in its pressurse will be
(A) 0.4%
16.
(D)
At a given volume and temperature the pressure of a gas
(A) 2500 C
14.
(C) 2
(B) 0.1 %
(C) 0.8%
(D) 0.2%
The product of the pressure and volume of an ideal gas is
(A) A constant
(B) Directly proportional to its temperature.
(C) Inversely proportional to its temperature.
(D) Approx. equal to the universal gas constant.
17.
o
o
At O C the density of a fixed mass of a gas divided by pressure is x. At 100 C, the ratio will
be
 273 
x
(B) 
 373 
(A) x
18.
(B) 206 kPa
(D) 212 kPa
(B) 17.26 ml
(C) 19.27 ml
(D) 192.7 ml
o
2g of O2 gas is taken at 27 C and pressure 76 cm Hg. Find out volume of gas (ln litre)
(A) 3.08
21.
(C) 200 kPa
The volume of a gas at 20 C is 200 ml. If the temperature is reduced to –20o C at constant
pressure, its volume will be.
(A) 172.6 ml
20.
 100 
x
(D) 
 273 
Air is pumped into an automobile tube upto a pressure of 200 kPa in the morning when the air
o
o
temperature is 22 C. During the day, temperature rises to 42 C and the tube expands by 2%
The pressure of the air in the tube at this temperature will be approximately.
(A) 209 kPa
19.
 373 
x
(C) 
 273 
(B) 44.2
(C) 1.53
(D) 2.44
1 mole of gas occupies a volume of 100 ml at 50 mm pressure. What is the volume occupied by
two moles of gas at 100 mm pressure and at same temperature
(A) 50 ml
(B) 200 ml
(C) 100 ml
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(D) 500 ml
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22.
23.
A partition divides a container, having insulated walls, into two
compartments, I and II. The same gas is filled the compartments.
The ratio of number of molecules in compartments I and II is.
(A) 6:1
(B) 1:6
(C) 4:1
P, V, T
2P, 2V, T
I
II
(D) 1:4
A cylinder contains 10 kg of gas at pressure of 107 N/m2. The quantity of gas taken out of the
cylinder, if final pressure is 2.5  10 6 Nm 2 , will be ______ (temperature of gas is constant)
(A) 5.2 kg
24.
(B) 3.7 kg
(D) 1 kg
o
The volume of a gas at pressure 21  104 Nm–2 and temperature 27 C is 83 Litres.
If R = 8.3 J mol–1K–1. Then the quantity of gas in g-mole will be
(A) 42
25.
(C) 7.5 kg
(B) 7
(C) 14
(D) 15
o
The pressure and temperature of an ideal gas in a closed vessel are 720 kPa and 40 C respectively.
If
1
4
th
of the gas is released from the vessel and the temperature of the remaning gas is raised
o
to 353 C, final pressure of the gas is
(A) 1440 kPa
26.
27.
(B) 2 P
o
o
(C) 851 C
o
(D)None of these
o
(C) 646 C
o
(D) 546 C
To double the volume of a given mass at an ideal gas at 270 C keeping the pressure constant one
must raise the temperature in degree centigrade
o
o
o
(B) 600
(C) 327
o
(D) 270
At constant temperature on incerasing the pressure of a gas 5% its volume will decrease by
(A) 5%
(B) 5.26%
(C) 4.76%
(D) 4.26%
o
o
At on 0 C pressure measured by barometer is 760 mm. what will be pressure at 100 C
(A) 780 mm
33.
o
(B) 651 C
(B) 182 C
(A) 54
32.
(D) 6 P
At what temperature volume of an ideal gas becomes triple
(A) 819 C
31.
(C) 4 P
Air is filled in a bottle at atmospheric pressure and it is corked at 35 C, If the cork can come
out at 3 atmospheric pressure then upto what temperature should the bottle be heated in order to
remove the cork.
o
30.
(D) 720 kPa
o
(A) 325.5 C
29.
(C) 1080 kPa
Suppose ideal gas equation follows VP 3 = constant, Initial temperature and volume of the gas are
T and V respectively. If gas expand to 27 V, then temperature will become
(D) T
(A) 9 T
(B) 27 T
(C) T 9
The temperature of a gas at pressure P and volume V is 270 C Keeping its volume constant if its
temperature is raised to 9270 C, then its pressure will be
(A) 3 P
28.
(B) 540 kPa
(B) 760 mm
(C) 730 mm
(D) None of these.
Hydrogen gas is filled in a ballon at 200 C. If temperature is made 400 C, pressure remaining the
same what fraction of haydrogen will come out
(A) 0.75
(B) 0.07
(C) 0.25
(D) 0.5
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34.
35.
36.
37.
38.
When the pressure on 1200 ml of a gas is increased from 70 cm to 120 cm of mercury at constant
temperature, the new volume of the gas will be
(A) 400 ml
(B) 600 ml
(C) 700 ml
(D) 500 ml
0
A gas at 27 C temperature and 30 atmospheric pressure is allowed to expand to the atmospheric
pressure if the volume becomes two times its initial volume, then the final temperature becomes
(A) 2730 C
(B) -1730 C
(C) 1730 C
(D) 1000 C
A gas at 270 C has a volume V and pressure P. On heating its pressure is doubled and volume
becomes three times. The resulting temperature of the gas will be
(A) 15270 C
(B) 6000 C
(C) 1620 C
(D) 18000 C
o
A perfect gas at 270 C is heated at constant pressure to 3270 C. If original volume of gas at 27 C
o
is V then volume at 327 C is
(A) 2 V
(B) V
(C) V 2
(D) 3 V
A vessel contains 1 mole of O2 gas (molar mass 32) at a temperature T. The pressure of the gas
is P. An identical vessel containing one mole of He gas (molar mass 4) at temperature 2T has a
pressure of
P
8
The pressure and temperature of two different gases P and T having the volumes V for each. They
are mixed keeping the same volume and temperature, the pressure of the mixture will be,
(A) 2 P
39.
(B) P
41.
42.
43.
(B)
2
E
3
3
E
2
E
2
The root mean square speed of hydrogen molecules of an ideal hydrogen kept in a gas chamber
at 00C is 3180 ms–1. The pressure on the hydrogen gas is (Density of hydrogen gas is
(A) P 
44.
(D)
P
(C) 4 P
(D) 2 P
2
Air is filled at 600 C in a vessel of open mouth. The vessel is heated to a temperature T so that
th
1
part of air escapes. Assuming the volume of the vessel remaining constant the value of T is.
4
(A) 3330 C
(B) 1710 C
(C) 4440 C
(D) 800 C
A gas is filled in a cylinder, its temperature is incresecd by 20% on kelvin scale and volume is
reduced by 10%. How much percentage of the gas will leak out
(A) 15%
(B) 25%
(C) 40%
(D) 30%
The pressure is exerted by the gas on the walls of the container because
(A) It sticks with the walls
(B) It is accelerated towards the walls
(C) It loses kinetic energy
(D) On collision with the walls there is a change in momentum
The relation between the gas pressure P and average kinetic energy per unit volume E is
(A) P
40.
(C) 8 P
(B) P 
(C) P  E
(D) P =
8.99  10 2 kg m 3 , 1 atm  1.01 105 Nm 2 )
(A) 1.0 atm
(B) 3.0 atm
(C) 2.0 atm
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(D) 1.5 atm
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45.
Gas at a pressure Po is contained in a vessel. If the masses of all the molecules are halved and
their speeds are doubled, the resulting pressure will be equal to
(A) 2 P0
46.
(B) 4 P0
(C)
P0
2
(D) P0
A cylinder of capacity 20 litres is filled with H 2 gas. The total average kinetic energy of translatory
motion of its molecules is 1.5  105 J. The pressure of hydrogen in the cylinder is
(A) 4  106 Nm–2
(B) 3  106 Nm–2
(C) 5  106 Nm–2
(D) 2  106 Nm–2
47.
The average kinetic energy per molecule of a gas at -230 C and 75 cm pressure is 5  10 14 erg for
H 2 . The mean kinetic energy per molecule of the O2 at 227 C and 150 cm pressure will be
48.
49.
50.
51.
52.
53.
54.
55.
(A) 80  10 14 erg
(B) 10  10 14 erg
(C) 20  10 14 erg
(D) 40  10 14 erg
The ratio of mean kinetic energy of hydrogen and oxygen at a given temperature is
(A) 1:8
(B) 1:4
(C) 1:16
(D) 1:1
The ratio of mean kinetic energy of hydrogen and nitrogen at temperature 300 K and 450 K
respectively is
(A) 2:3
(B) 3:2
(C) 4:9
(D) 2:2
Pressure of an ideal gas is increased by keeping temperature constant what is the effect on kinetic
energy of molecules.
(A) Decrease
(B) Increase
(C) No change
(D) Can`t be determined
A sealed container with negligible co-efficient of volumetric expansion contains helium (a monoatomic
gas) when it is heated from 200 K to 600 K, the averagy K.E. of helium atom is
(A) Halved
(B) Doubled
(C) Unchanged (D) Increased by factor 2
The mean kinetic energy of a gas at 300 K is 100J. mean energy of the gas at 450 K is equal to
(A) 100 J
(B) 150 J
(C) 3000 J
(D) 450 J
At what temperature is the kinetic energy of a gas molecule double that of its value at 27 C
(A) 540 C
(B) 1080 C
(C) 3270 C
(D) 3000 C
The average kinetic energy of a gas molecule at 270 C is 6.21  10–21 J. Its average kinetic energy
at 2270 C will be
(A) 5.22  10 21 J
(B) 11.35 10 21 J
(C) 52.2  10 21 J
(D) 12.42  1021 J
The average translational energy and rms speed of molecules in sample of oxygen gas at 300 K
are 6.21 1021 J and 484 m s respectively. The corresponding values at 600 K are nearly
(assuming ideal gas behaviour)
(A) 6.21 1021 J , 968 m s
(B) 12.42  1021 J , 684 m s
(C) 12.42  1021 J , 968 m s
(D) 8.78  1021 J , 684 m s
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56.
57.
58.
The average translational kinetic energy of O2 (molar mass 32) molecules at a particular temperature
is 0.068 eV. The translational kinetic energy of N2 (molar mass 28) molecules in eV at the same
temperature is
(A) 0.003 eV
(B) 0.068 eV
(C) 0.056 eV
(D) 0.678 eV
At O K which of the follwing properties of a gas will be zero
(A) Kinetic energy
(B) Density
(C) Potential energy (D) Vibrational energy
The kinetic energy of one mole gas at 300 K temperatue is E. At 400 K temperature kinetic enrgy
is E'. The value of
16
4
(C) 1.33
(D)
9
3
The average kinetic energy of hydrogen molecules at 300 K is E. At the same temperature the
average kinetic energy of oxygen molecules will be
(A) 2
59.
60.
(B)
(A) E 16
(B) E
(C) 4 E
(D) E 4
The temperature at which the average translational kinetic energy of a molecule is equal to the
energy gained by an electron accelerating from rest through a potential differencc of 1 volt is
(A) 4.6 103 K
61.
62.
63.
64.
66.
(B) 7.7 103 K
(C) 11.6 103 K
(D) 23.2 103 K
At a given temperature the rms velocity of molecules of the gas is
(A) Proportional to molecular weight
(B) Inversely proportional to molecular weight
(C) Inversely proportional to square root of molecular weight
(D) Proportional to square of molecular weight
According to the kinetic theroy of gases the r.m.s velocity of gas molecules is directly proportional
to
1
(B) T
(A) T 2
(C) T
(D)
T
The speeds of 5 molecules of a gas (in arbitrary units) are as follws: 2, 3, 4, 5, 6, The root mean
squre speed for these molecules is
(A) 4.24
(B) 2.91
(C) 4.0
(D) 3.52
o
To what temperature should the hydrogen at room temperature (27 C) be heated at constant
pressuse so that the rms velocity of its molecule becomes double of its previous value
(A) 927 C
65.
E'
is
E
(B) 600 C
(C) 108 C
(D) 1200 C
Root mean square velocity of a molecule is  at pressure P. If pressure is increased two times,
then the rms velocity becomes
(A) 3
(B) 2
The rms speed of gas molecules is given by
(A) 2.5
Mo
RT
RT
(C) 0.5
(D) 
RT
(B) 2.5 M
o
(C) 1.73 M
o
(D) 1.73
300
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Mo
RT
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o
67.
A sample of gas is at O C. To what temperature it must be raised in order to double the rms
speed of molecule.
o
o
o
o
(A) 270 C
(B) 819 C
(C) 100 C
(D) 1090 C
68.
If the ratio of vapour density for hydrogen and oxygen is
ratio of their rms velocities will be
(A) 4:1
(B) 1:16
69.
1
, then under constant pressure the
16
(C) 16:1
(D) 1:4
The molecules of a given mass of a gas have a rms velocity of 200 m s at 27 C and
1.0  105 Nm 2 pressure when the temperature is 127 C and pressure is 0.5  10 5 Nm 2 , the
rms velocity in m s will be
(A) 100 2
70.
(B)
100 2
3
(C)
400
3
(D) None of these
If the molecular weight of two gases are M1 and M 2 , then at a given temperature the ratio of
root mean square velocity 1 and  2 will be
71.
72.
(A)
M1
M2
(B)
M2
M1
(C)
M1  M 2
M1  M 2
(D)
M1  M 2
M1  M 2
To what temperature should the hydrogen at 327 C cooled at constant pressure, so that the root
mean square velocity of its molcules become half of its previous value
(B) 123 C
(C) 0 C
(D) 123 C
(A) 100 C
At what temperature is the root mean square velocity of gaseous hydrogen molecules equal to that
of oxygen molecules at 47 C ?
(A) -73 K
73.
(D) 3 K
th
(B) 0 C
(C) O K
(D) 100 C
At room temperature ( 27 C ), the rms speed of the molecules of certain diatomic gas is found to
be 1930 m/s. The gas is
(A) O2
75.
(C) 20 K
The root mean square velocity of the molecules in a sample of helium is 57 that of the molecules
in a sample of hydrogen. If the temperature of hydorgen sample is 00C, then the temperature of
the helium sample is about
(A) 273 C
74.
(B) 80 K
(B) C 2
(C) H2
(D) F2
If three molecules have velocities 0.5, 1 and 2 the ratio of rms speed and average speed is (The
velocities are in km/s)
(A) 0.134
(B) 1.34
(C) 1.134
(D) 13.4
301
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76.
77.
78.
79.
80.
81.
At what temperature pressure remaining constant will the rms speed of a gas molecules increase
by 10% of the rms speed at NTP?
(A) 57.3 K
(B) 57.30 C
(C) 557.3 K
(D) -57.30 C
When temperature of an ideal gas is increased from 27o C to 227o C, its rms speed changed from
400 ms–1 to Vs. The Vs is
(A) 516 ms–1
(B) 746 ms–1
(C) 310 ms–1
(D) 450 ms–1
At what temperature the molecules of nitrogen will have the same rms. velocity as the molecules
of Oxygen at 127oC.
(A) 2730 C
(B) 3500 C
(C) 770 C
(D) 4570 C
The temperature of an ideal gas is increased from 270 C to 1270 C, then percentage increase in
rms is
(A) 33%
(B) 11%
(C) 15.5%
(D) 37%
Let A and B the two gases and given : TA  4 M A . Where T is the temperature and M is
TB
MB
A
molecular mass. If A and B are the r.m.s speed, then the ratio  will be equal to __________
B
(A) 2
(B) 4
(C) 0.5
(D) 1
-1
The rms. speed of the molecules of a gas in a vessel is 400 ms . If half of the gas leaks out,
at constant temperature, the r.m.s speed of the remaining molecules will be
(A) 800 ms-1
82.
83.
84.
85.
86.
87.
(B) 200 ms-1
(C) 400 2 ms-1 (D) 400 ms-1
At which temperature the velocity of 02 molecules will be equal to the rms velocity of N2 molecules
at 00 C
(A) 930 C
(B) 400 C
(C) 390 C
(D) can not be calculated.
The rms speed of the molecules of a gas at a pressure 105 Pa and temperature 00 C is
0.5 km/s. If the pressure is kept constant but temperature is raised to 8190 C, the rms speed
becomes
(A) 1.5 kms-1
(B) 2 kms-1
(C) 1 kms-1
(D) 5 kms-1
The root mean square velocity of a gas molecule of mass m at a given temperature is proportional
to
(A) m0
(B) m-1/2
(C) m1/2
(D) m
The ratio of the vapour densities of two gases at a given temperature is 9:8, The ratio of the rms
velocities of their molecule is
(A) 3 : 2 2
(B) 2 2 : 3
(C) 9:8
(D) 8:9
At what temperature, pressure remaining unchanged, will the rms velocity of a gas be half its value
at OoC ?
(A) 204.75 K
(B) 204.750 C
(C) -204.75 K (D) -204.750 C
The rms velocity of gas molecules is 300 ms-1. The rms velocity of molecules of gas with twice
the molecular weight and half the absolute temperature is
(A) 300 ms-1
(B) 150 ms-1
(C) 600 ms-1
(D) 75 ms-1
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88.
89.
90.
Calculate the temperature at which rms velocity of S02 molecules is the same as that of O2
molecules at 270 C. Molecular weights of Oxygen and SO2 are 32 g and 64 g respectively
(A) 3270 C
(B) 327 K
(C) 1270 C
(D) 2270 C
For a gas, the rms speed at 800 K is
(A) Four times the value at 200 K
(B) Twice the value at 200 K
(C) Half the value at 200 K
(D) same as at 200 K
A mixture of 2 moles of helium gas (atomic mass = 4 amu), and 1 mole of argon gas (atomic mass
= 40 amu) is kept at 300 K in a container. The ratio of the rms speeds
91.
92.
93.
94.
95.
(A) 0.45
(B) 2.24
(C) 3.16
(D) 0.32
0
0
The temperature of an ideal gas is increased from 27 C to 927 C. The root mean square speed
of its molecules becomes
(A) Four times
(B) One-fourth (C) Half
(D) Twice
At a given temperature the root mean square velocities of Oxygen and hydrogen molecules are in
the ratio
(A) 1:4
(B) 1:16
(C) 16:1
(D) 4:1
If mass of He atom is 4 times that of hydrogen atom, then rms speed of the is
(A) Two times of H  rms speed.
(B) Four times of H  rms speed.
(C) Same as of H  rms speed.
(D) half of H  rms speed.
At temperature T, the rms speed of helium molecules is the same as rms speed of hydrogen
mdecules at normal temperature and pressure. The value of T is
(A) 5460 C
(B) 00 C
(C) 2730 C
(D) 136.50 C
The root mean square speed of hydrogen molecules at 300 K is 1930 m/s. Then the root mean
square speed of Oxygen molecules at 900 K will be
(A) 836 m/s
96.
97.
98.
vrms (helium)
is
vrms (argon)
(B) 643 m/s
(C) 1930 3 m/s
(D)
1930
m/s
3
If rms speed of a gas is rms = 1840 m/s and its density  = 8.99  10-2 kg/m3 , the pressure of
the gas will be
(A) 1.01  103 Nm–2
(B) 1.01  105 Nm–2 (C) 1.01  107 Nm–2 (D) 1.01 Nm–2
When the temperature of a gas is raised from 270 C to 900 C , the percentage increase in the
rms velocity of the molecules will be
(A) 15%
(B) 17.5%
(C) 10%
(D) 20%
The rms speed of a gas at a certain temperature is 2 times than that of the Oxygen molecule
at that temperature, the gas is_____
(A) SO2
(B) CH4
(C) H2
(D) He
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99.
The temperature at which the rms speed of hydrogen molecules is equal to escape velocity on earth
surface will be
(A) 5030 K
(B) 10063 K
(C) 1060 K
(D) 8270 K
100. What is the meanfree path and collision frequency of a nitrogen molecule in a cylinder containing

nitrogen at 2 atm and temperature 17oC ? Take the radius of nitrogen molecule to be 1A .
M olecular mass of nitrogen = 28 , k B = 1.38  10–23 JK–1, 1 atm = 1.013  105 Nm–2
(A) 2.2 10 7 m, 2.58  10 9
(B) 1.110 7 m, 4.58  108
(C) 1.110 7 m, 4.58 10 9
(D) 2.2 10 7 m, 3.58  109
0
101. The radius of a molecule of Argon gas is 1.78 A . Find the mean free path of molecules of Argon
at 00 C temperature and 1 atm pressure.
k B  1.38  10 23 JK 1
(A) 6.65  10 8 m
(B) 3.65  10 7 m
(C) 6.65  10 7 m
(D) 3.65  10 8 m
102. A monoatomic gas molecule has
(A) Three degrees of freedom
(B) Five degrees of freedom
(C) Six degrees of freedom
(D) Four degrees of freedom
103. A diatomic molecule has how many degrees of freedom (For rigid rotator)
(A) 4
(B) 3
(C) 6
(D) 5
104. The degrees of freedom for triatomic gas is ______ (At room temperature)
(A) 8
(B) 6
(C) 4
(D) 2
105. If the degrees of freedom of a gas are f, then the ratio of two specific heats
Cp
Cv
is given by
2
2
1
3
1
(B) 1 
(C)
(D) 1 
f
f
f
f
106. A diatomic gas molecule has translational, rotational and vibrational degrees of freedom. The
(A) 1 
Cp
Cv
is
(A) 1.29
(B) 1.33
(C) 1.4
107. The value of Cv for one mole of neon gas is
(A)
3
R
2
(B)
7
R
2
(C)
1
R
2
(D) 1.67
(D)
5
R
2
108. The relation between two specific heats of a gas is
(A) CV  CP  R J
(B) CP  CV  J
(C) CP  C V  R J
(D) CV  CP  J
304
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109. The molar specific heat at constant pressure for a monoatomic gas is
(A)
3
R
2
110. For a gas
(B)
5
R
2
(C) 4 R
(D)
7
R
2
R
 0.67 . This gas is made up of molecules which are
CV
(A) Diatomic
(B) monoatomic
(C) polyatomic
(D) mixture of diatomic and polyatomic molecules
111. The specific heat of an ideal gas is
(A) Proportional to T2
(B) Proportional to T3
(C) Proportional to T
(D) Independent of T
112. The specific heats at constant pressure is greater than that of the same gas at constant volume
because (A) At constant volume work is done in expanding the gas.
(B) At constant pressure work is done in expanding the gas.
(C) The molecular vibration increases more at constant pressure.
(D) The molecular attraction increases more at constant pressure.
 7
is mixed with one mole of diatomic gas     .
3

What is  for the mixture?  denotes the ratio of specific heat at constant pressure to that at
constant volume.
113. One mole of ideal monoatomic gas
(A)
114.
35
23
(B)
23
15
(C)
3
2
(D)
4
3
For a gas if ratio of specific heats at constant pressure and volume is  , then value of degrees
of freedom is
2
(A)   1
115.
5 

  3 


(B)
25
  1
2
3  1
(C) 2   1
The molar specific heat at constant pressure of an ideal gas is
(D)
7
R . The ratio of specific heat
2
at constant pressure to that ratio at constant volume is
(A)
116.
5
7
(B)
For a gas  
9
7
(C)
8
7
9
  1
2
(D)
7
5
7
, the gas may probably be
5
(A) Neon
(B) Argon
(C) Helium
(D) Hydrogen
117. From the following P - T graph, what inference can be drawn
(A ) V 2< V1
(B) V2= V1
(C) V 2>V1
(D) none of these
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118. The figure shows the volume V versus temperature T graphs for
a certain mass of a perfect gas at two constant pressure of
P1 and P2. What inference can you draw from the graphs.
(A) P1< P2
(B) P1>P2
(C) P1= P2
(D) No inference can be drawn due to insufficient information.
119. Which one the following graphs represents the behaviour of an ideal gas
(A)
(B)
(C)
120. Under constant temperature, graph between p and 1 V is
(A) Hyperbola
(B) Circle
(C) Parabola
(D)
(D) Straight line
Direction:(a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.
(b) If both Assertion and Reason are true, but Reason is not correct explanation of the Assertion.
(c) If Assertion is true; the Reason is false.
(d) If Assertion is false, but the reason is true.
121. Assertion : 300 cc of a gas at 270 C is cooled at -30 C at constant pressure. The final volume
of the gas would be 270 cc
Reason
V2 T2
: This is as per charle's law V  T
1
1
(A) a
(B) b
(C) c
(D) d
-8
122. Assertion : The time of collision of molecules is of the order of 10 s, which is very very small
compared to the time between two successive collisions.
Reason
: This is an experimental fact.
(A) a
(B) b
(C) c
(D) d
123. Assertion : Mean free path of gas varies inversly as density of the gas.
Reason
: Mean free path varies inversely as pressure of the gas.
(A) a
(B) b
(C) c
(D) d
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KEY NOTE
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
1 B
2 C
3 B
C
4
A
5 C
6 A
7 A
8 C
B
9 D
10 D
11 C
12A
A
13
B
14 B
15A
16A
17 C
C
18
D
19 C
20 B
21 C
22
23
24
25
B
C
B
C
A
C
A
A
C
B
D
D
C
A
A
B
B
A
A
C
C
D
C
B
C
26
26
27
27
28
28
29
29
30
30
31
31
32
33
32
34
33
35
34
36
35
37
38
36
39
37
40
38
41
39
42
40
43
44
41
45
42
46
43
47
44
48
45
49
50
46
47
48
49
50
A
C
B
D
C
C
D
B
C
B
A
A
A
D
B
B
D
A
B
C
B
D
A
A
C
A51
C52
B53
D
54
C
C55
D56
B57
C
58
B
A59
A60
A61
D62
B
63
B
D64
A65
B66
A67
C
68
B
D69
A70
C71
72
73
74
75
B 51
B 52
C 53
54
D
55
B 56
B 57
A 58
C 59
60
B 61
B 62
C 63
B 64
65
A
66
A 67
D 68
C 69
B 70
71
A
72
C 73
B 74
D 75
C
B
C
C
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
BB
AB
CC
D
C
B
AB
DA
CC
CB
B
BC
BB
DA
BA
D
A
C
BB
CA
DC
AB
D
D
C
CB
AC
BC
C
B
B
C
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
A
76
77
A
78
D
79
B
80
D
81
B
82
83
A
84
C
85
B
86
B
87
88
D
89
90
C
91
A
92
D
93
94
D
95
C
96
B
97
C
98
D
99
100
A
B
A
C
C
A
D
C
C
B
B
D
B
A
B
C
D
A
D
C
A
B
C
B
B
C
A
B
307
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101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
A
A
D
B
C
B
A
C
B
B
D
B
C
A
D
D
C
B
C
D
A
A
B
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Hint / Solution
1.
M
PV = µRT  PV  M
(B)

PM o M


RT
V
 
2.
RT
O
(C)
PM o P  m  N A
Pm
Pm



 R 
RT
RT
k BT

T
 NA 
PV  RT  P
T
V
If T and V both doubled, then pressure remains same
3.
(B)
PV  μRT  PV α T
If P and V doubled, then T becomes four times,
4.
(C)
PV  µRT  µ 
PV
RT
 Mass of  litre nitrogen = µ  M o
5.
(A)
Gas equation for N molecules, PV = NkBT
 N 
6.
(C)
PV
k BT
PV  µRT  P 
µRT
V
For same µ, R and V, P  T
7.
(A)
molecular weight of oxygen = 32 (g)
number of moles in 5 g of oxygen =

Equation of state is PV = µRT
 PV 
8.
(A)
5
32
5
RT
32
Total number of moles is conserved,

PV
PV
PV
1 1
 2 2 
RT
RT
RT
1 100
0 .5  5 0
P 1000
+
=
( 1 L = 100 m  )
RT
RT
RT
 P  0.125 atm
308

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9.
(C)
The ideal gas equation is
PV  µRT
PV  RT  µ  1


R
 PV  k B N AT 
 kB 
NA


PV

 N A  Avogadro's number..
k BT
10.
(B).
PV  µRT
 M 
 PV  
 RT
 Mo 
 M  RT
  
 V  Mo
 RT
Mo

11.
(D)
 M

   
 V

P
R

= const.
T M o
PV  µRT
M
 P    µRT
  
M  M 
P
 P 
 RT   
T
    Mo 
12.
(D)
PV  µRT
M 
 PV  
 RT  P  M
 Mo 
13.
(C). PV  µRT
 P T

14.
( V, T  Constant )
( Closed vessel i.e. volume is constant)
P1 T1

P2 T2
(A). PV  µRT = Constant ( temp. is const.)
 P1V1  P2V2
 95 
 PV
1 1  P2 
V1
 100 
 V2  95% V1 
309
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 P2  1.0526 P1
 P1  0.0526 P1
 P1  5.26% P1
15.
 Pressure 5.26% increases.
(A) Closed Vessel . i.e. volume remains constant.
From, PV  µRT
P α T 

P2 T2

P1 T1
P2  P1 T2  T1

P1
T1
16.
(B)
 PV  T (  µ,R  Constant)
17.
(B)
PV  µRT
 M 


 Mo 
RT
M Mo

PV RT
density
Mo

=
P
RT

 density 


 P At 0 C
 density 
 P 

 At 100 C
M
 x...........................(i)
R  273
=
=
M
.........................(ii)
 R  373
 density 
 273 

x
  P 

 At 100 C  373 
18.
(A)
PV
=µR = constant  P1V1  P2 V2
T
T1
T2
19.
(A)
PV=µRT
since P is const. V  T

20.
(C)
V1 T1

V2 T2
PV= µRT
 M 
=
 RT
 Mo 
V=
MRT
M oP
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21. (C)
PV= µRT

22.
(D)
P1V1
µ
 1
P2 V2
µ2
PV  Nk B T
Now, N ' 
23.
(C)
( T const.)
 N
PV
k BT
 2P  2V 
 PV 
 4
4N
 k BT 
k BT

N 1

N' 4
PV=µRT
 M 
 PV = 
 RT  P  M
 Mo 

( V, R, T - Constant)
P1 M1
107
10



 M2 = 2.5 kg
6
P2 M 2
2.5  10
M2
 Hence mass of the gas taken out of the cylinder = 10 – 2.5 = 7.5 kg
PV
RT
24.
(B)
PV = µRT   =
25.
(C)
 M 
PV = µRT  PV = 
 RT  P  MT
 Mo 
26.
(A)
VP3 = constant = k  P 3 
k
V
k
V
k V
 V
2
 V1 
Hence  
 V2 

1 T

9 T2
3
2
3
2
3

1
P2  M 2   T2 

 
P1  M1   T1 
1
1
 P 1
V
V 3
 P
PV = µRT 

1
3
V  µRT
3
 µRT
µRT
k
2
T
 1
T2
 V 


 27 V 
3

T
T2
 T2  9 T
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P1 T1
Using Gay-lussac's law P  T
2
2
27.
(C)
28.
(B) At constant volume
P1
T1
29.

P2
T2
 P 
T2   2 
 P1 

T1
(D) At constant Pressure V  T

V2 T2

V1 T1
30.
(C)
V  T
31.
(C)
P
V
 T2   2
 V1

 T1

V1 T1

V2 T2
1
V
P
100
100
V1  0.9524 V1
 2 1 
 V2 
V
V1 P2
105
105
 V2  1  0.0476  V1
 V1  0.0476 V1
 V1  4.76% V1
32.
(D)
PT 
P2 T2
T

 P2  P1 2
P1 T1
T1
33.
(B)
V T 
V2 T2
V  V1 T2  T1

 2

V1 T1
V1
T1

34.
(C)
313  293
V  273  40    273  20 

 0.07

293
V
273  20
At constant Pressure PV = constant
 P1V1  P2 V2 
P1 V2

P2 V1
PV 
 T2   2 2  T1
 P1V1 
35.
(B)
P1V1 P2 V2

T1
T2
36.
(A)
PV
P1V1
PV
 2 2  T2  2 2 T1
Pl V1
T1
T2
37.
(A)
VT 
V1 T1

V2 T2
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38.
(A)
PV = µRT  P=µT ( V and R = constant )
P2 µ 2 T2

P1 µ1T1

39.
(D)
PV= µRT  µ1 
 P '
40.
(B)
 µ1  µ 2  RT

V
PV
PV
and µ 2 
RT
RT
2 PV RT

 2P
RT
V
For open mouth vessel, pressure is constant.
volume is also given constant.
Hence from  PV  µRT
PV 


41.
(B)
M
1
RT  T 
Mo
M

T1 M 2

T2 M1
1 th
part escapes, so remaining mass in the vessel is
4
3
M
3
273  60 4 1
M 2  M1 
 T  444 K  1710 C

4
T
M1
Let initial conditions = V, T
final conditions = V', T'
By Charle's law, V  T ( P remains constant )
V
V'
V
V'



 V '  1.2 V
T
T'
T 1.2 T
But as per question, volume is reduced by 10%
means  V' = 0.9 V
so percentage of volume leaked out

1.2  0.9  V
1.2 V
42.
(D)
Pressure P 
43.
(A)
P

44.
(B)
 100 %  25%
F 1 P
( P = change in momentum)

A A t
2
( Energy per unit volume )
3
2
3
E
V
 rms 

2
E
3
3P
2 
 P  rms

3
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3P
3PV


M
45.
(A)
 rms 
46.
(C)
E
47.
(B)
The average kinetic energy
E
  rms 
P
M
1

2

P1 M 2

P2 M1
3
2E
PV  P 
2
3V
3
E
T
k BT  1  1
2
E 2 T2
48.
(D)
Kinetic energy is a function of temperature.
49.
(A)
ET
50.
(C)
51.
(B)
Kinetic energy of ideal gas depends only on its temperature. Hence, it remains constant
whether pressure is increased or decreased.
Kinetic energy is directly proportional to temperature. Hence if temperature is doubled,
kinetic energy will also be doubled.
52.
(B)
53.
(C)
ET 
E1 T1

E 2 T2
54.
(D)
ET 
E1 T1

E 2 T2
55.
(B)
Average translational K.E. of a molecule is =

E1 T1

E 2 T2
Average kinetic energy  Temperature 
E1 T1

E 2 T2
At 300 K, average K.E.
= 6.21  10 21 J
At 600 K average K.E.
= 2  6.21  10 21
3
k BT
2
= 12.42  1021 J
We know that rms =
3k BT
m
At 300 K, rms = 484 ms–1
At 600 K, rms =
56.
57.
2  484  684 ms 1
3
k BT
2
(B)
Average translational K.E. of a molecules 
(A)
(Where, kB = Boltzmann's constant )
This is same, for all gases at same temperature.
At 0 K Kinetic energy is zero.
314
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3
E' T'
RT  E  T 

2
E T
58.
(C)
E
59.
(B)
E T
60.
(B)
2 eV 2
1.6  10 19
3
 
 7.7  103 K
k B T  1 eV  T 
23
3 kB 3
1.38  10
2
61.
(C)
rms =
3RT
M o  rms 
62.
(B)
rms 
T
2
2
2
1
Mo
2
2
63.
(C)
rms =
1   2   3   4   5
5
64.
(A)
rms 
  rms 2

T    rms 
1
65.
(D)
rms velocity does not depend on pressure.
66.
(C)
rms =
3RT
Mo
67.
(B)
rms 
T , To double the rms speed temperature should be made four times i.e.

3
= 4.24
T2
T1
RT
Mo
 1.73
RT
Mo
 T2  4T1
3P

 1 

2
2

1
16
 4 :1
1
68.
(A)
vrms 
69.
(C)
rms velocity doesn't depend on pressure, it depends upon temperature only.
 rms =
3RT
  rms  T  T   2 rms
Mo

1
T1

2
T2
70.
(B)
 rms =
3RT
1
1

 1 
 1 
and  2 
Mo
M1
M2
2
71.
(D)
 rms =
3RT
  rms  T  T   2rms
Mo
72.
(C)
 rms 
3RT
Mo
M2
M1
 T  M o (  rms , R  constant)
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73.
(B)
3RT
T
  rms 
Mo
Mo
 rms 

THe  M H 2
 He 5
 
 H2 7
TH 2 M He
 THe 
25 4
  273
49 2
 273 K  00 C
74.
(C)
3 RT
Mo
 rms 
 Mo 
3 RT 3  8.3  300

 2  103 kg  2 g
2
 2 rms
1920 
Gas is hydrogen.
75.
(C)
rms speed, rms =
1   2   3
3
average speed,  
76.
(B)
As
 1.1 
  rms t
  rms O
(A)
 rms  T
78.
(C)
 rms 
79.
TN2
TO 2
(C)

T
and rms speed increases by 10%
T0
273  t
273  t
2
 1.1  1.21
or
273
273
77.


12   22  32
3
2
T2

1
T1
3RT
 T  Mo
Mo
 M O N
 M O O
 rms 

 t  273 1.21  1  57.30 C
(  rms , R  constant)
2
2
3RT
Mo
% increase in  rms 
3RT2
3 RT1

Mo
Mo
3RT1
Mo
 100% 
20  17.32
100%  15.5%
17.32
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80.
(A)
TA
T
4 B
MA
MB


TA
MA
 2
3 RTA
3 RTB
 2
MA
MB
  A  2 B 
A
2
B
81.
(D)
Since temperature is constant.
so vrms remains same.
82.
(C)
 rms =

3 RT
 T  Mo
Mo
TO2

TN2
 M o O
 Mo N
(  rms , R  Const. )
2
2
  rms 1
  rms 2

T1
T2
83.
(C)
 rms  T 
84.
(B)
 rms 
85.
(B)
At a given temperature  rms 
86.
(D)
 rms  T
  rms 2
  rms 1


87.
TB
MB
3k BT
-1
  rms  m 2
m
and   rms  2 
T 1

T0 2


1

1
  rms 1
2
273  t
273  0

1
2
273  t 1
273
 t
 273 = 68.25-273 = -204.750 C
273
4
4
(B)
 rms 
3 RT
Mo
  rms 1 
3 RT
;
Mo
  rms 2 
3R  T2 
3RT

4M o
2M o

1 3RT
2 4M o

 vrms 1
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2
= 150 ms–1
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88.
(A)
3 RT
Mo
 rms 

3RT
 rms =
=

Mo

89.
(B)
28×10-3


  rms 1    rms 2
here

 3 8.314 290  508.24 ms1 
TO 2
 M o O

2
TSO2
 Mo SO

 rms  T
90.
(C)
  rms He
  rms Ar
91.
(D)
 rms  T 
92.
(A)
 rms 

1
Mo
300 TSO2
0

 Tso 2  600 K  327 C
32
64

2
v1
T
 1
T2
v2
3RT
 Mo He
3RT
 M o Ar
40
 10
4

2
T2

1
T1
So
 rms O
  rms H
2
=
2
 M o H
 M o O
2
2
93.
(D)
 rms 
1

mH
 He 
m
m He
H
94.
(C)
rms =
 M o He
T
3RT
 T  M o  He =
TH  M o H
M0
95.
(A)
rms =
3 RT
Mo
96.
(B)
 rms =
3P

 3.16

 rms O
  rms H
2
2
=
TO2 ×  M o H
2
TH2 ×  M o  O
2
2
or P =   rms
3
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97.
(C)
3RT
Mo
 rms =

2
=
1
T2
=
T1
273+90
273+27
=1.1
 2 
 % increase =   1 ×100% = 0.1×100% =10%
 1 
98.
(B)
 rms 
1
Mo

1
=
2
 M o 2
 M o 1
 M o 2
1
=
2

  M o  2 = 16
32
 Hence the gas is CH 4
99.
(B)
Escape velocity from the earth's surface is 11.2 kms–1
2
So, rms = Vescape
100.
(C)  =
k BT
2πPd 2
 escape   Mo
3RT

T
Mo
3R
1.38×10   290
=
 2  3.14   2.026×10  2×10 
-23
-5
-10
Collision Frequency = no. of collision per second 

  rms =


101. (A)  
1
=
2πnd 2
2
= 1.11 10-7 m
508.24
 rms
=
= 4.58×109
-7
1.11×10

3RT
=
Mo
 38.314 290   508.24 ms1 
28×10-3


k BT
 6.65  108 m
2
2 πPd
102. (A) A monoatomic gas molecule has only three translational degrees of freedom.
103. (D) A diatomic molecule has three translational and two rotational degrees of freedom. Hence
total degrees of freedom, f = 3+2 = 5
104. (B) For a triatomic gas f = 6 ( 3 translation + 3 rotational )
105.
(C)
Cp
2
   1
Cv
f
106. (B) Degrees of freedom = 3 ( translatory ) + 2 ( rotatory ) + 1 ( vibratory ) = 6
Cp
2
2
1
   1
= 1 = 1
Cv
f
6
3

4
 1.33
3
107. (A) Neon gas is mono atomic and for mono atomic gases C v  3 R
2
108.
(C) When Cp and Cv are given with caloric and R with Joule then Cp – Cv =
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R
J
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109. (B) C p  C v  R
110.
(B) C v =
 Cp  R  Cv  R 
f
3
R = R+ R  f = 3
2
2
5
 R
2
R
3
= 1.5 R =
R
0.67
2
This is the value for mono atomic gases
111. According to the equilibrium theorem, the molar heat capacities should be independent of temperature
How ever, variations in Cv and Cp are observed as the temperature changes. At very high
temperatures, vibrations are also inportant and that affects the values of Cv and Cp for diatomic
and poly atomic gases. Here in the question according to given information (D) may be correct
answer.
112. (B)
1 53
1 75
µ11 µ 2  2


 53  1 75  1  3  1.5
1  1  2  1



113. (C) mix
µ1
µ
1
1
2
 2
 7
5
1  1
 2 1

1

1
3  5 
  
2
2
f
1
2
  1   
f 
f
f
2  1
 1
114.
(A)
115.
(D) molar specific heat at constant pressure, Cp 
Since C p  C v  R  C v  Cp  R 
116.
117.
118.
(D)  
7
R
2
7
5
R R  R
2
2
7
for a diatomic gas.
5
T T
(C) As 2  1  tan 2 > tan 1      
 P  2  P 1
T
Also from PV  µRT ,
 V  V2  V1
P
V
V
(B) As 1  2  tan 1 < tan 2      
 T 1
 T 2
1
1
V 1
Hence  P    P   P1  P2

 1  2
T P
119. (C) For an ideal gas PV = constant
i.e PV doesn't vary with V
1
120. (D) At constant temperature, PV = constant  P 
V
from PV = µRT ;
320
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