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 Page 1


Solved Examples on Kinetic Theory of 
Gases 
JEE Mains 
Single Correct 
Q1. The root mean square speed of hydrogen molecules of an ideal hydrogen gas kept in a gas 
chamber at ?? °
?? is ???????? ?? /?? . The pressure on the hydrogen gas is 
(Density of hydrogen gas is ?? . ???? × ????
-?? ???? /?? ?? , ?? atmosphere = ?? . ???? × ????
?? ?? /?? ?? ) 
(a) ?? . ?? ?????? 
(b) ?? . ?? ?????? 
(c) ?? . ?? ?????? 
(d) ?? . ?? ?????? 
Ans : (d) As ?? =
1
3
?? ?? rms
2
=
1
3
( 8.99 × 10
-2
)× ( 3180 )
2
= 3.03 × 10
5
 N/m
2
= 3.0 atm 
Q2. A flask contains ????
-?? ?? ?? gas. At a temperature, the number of molecules of oxygen are 
?? . ?? × ????
????
. The mass of an oxygen molecule is ?? . ?? × ????
-????
 ???? and at that temperature the ?????? 
velocity of molecules is ?????? ?? /?? . The pressure in ?? /?? ?? of the gas in the flask is 
(a) ?? . ???? × ????
?? 
(b) ?? . ???? × ????
?? 
(c) ???? . ???? × ????
?? 
(d) ???? . ???? × ????
?? 
Ans : (a) ?? = 10
-3
 m
3
, N = 3.0 × 10
22
, m= 5.3 × 10
-26
 kg , ?? ?????? = 400 m /s 
?? =
1
3
????
?? ?? ?????? 2
=
1
3
×
5.3 × 10
-26
× 3.0 × 10
22
10
-3
( 400)
2
= 8.48 × 10
4
 N/m
2
. 
Q3. A flask is filled with ???????? of an ideal gas at ????
°
?? and its temperature is raised to ????
°
?? . The 
mass of the gas that has to be released to maintain the temperature of the gas in the flask at ????
°
?? 
and the pressure remaining the same is 
(a) ?? . ?? ?? 
(b) ?? . ?? ?? 
(c) ?? . ?? ?? 
(d) ?? . ?? ?? 
Ans: (d) ???? ? Mass of gas × Temperature In this problem pressure and volume remains constant so 
?? 1
?? 1
= ?? 2
?? 2
= constant 
?
?? 2
?? 1
=
?? 1
?? 2
=
( 27 + 273)
( 52 + 273)
=
300
325
=
12
13
? ?? 2
= ?? 1
×
12
13
= 13 ×
12
13
gm= 12gm 
i.e. the mass of gas released from the flask = 13gm- 12gm= 1gm . 
Page 2


Solved Examples on Kinetic Theory of 
Gases 
JEE Mains 
Single Correct 
Q1. The root mean square speed of hydrogen molecules of an ideal hydrogen gas kept in a gas 
chamber at ?? °
?? is ???????? ?? /?? . The pressure on the hydrogen gas is 
(Density of hydrogen gas is ?? . ???? × ????
-?? ???? /?? ?? , ?? atmosphere = ?? . ???? × ????
?? ?? /?? ?? ) 
(a) ?? . ?? ?????? 
(b) ?? . ?? ?????? 
(c) ?? . ?? ?????? 
(d) ?? . ?? ?????? 
Ans : (d) As ?? =
1
3
?? ?? rms
2
=
1
3
( 8.99 × 10
-2
)× ( 3180 )
2
= 3.03 × 10
5
 N/m
2
= 3.0 atm 
Q2. A flask contains ????
-?? ?? ?? gas. At a temperature, the number of molecules of oxygen are 
?? . ?? × ????
????
. The mass of an oxygen molecule is ?? . ?? × ????
-????
 ???? and at that temperature the ?????? 
velocity of molecules is ?????? ?? /?? . The pressure in ?? /?? ?? of the gas in the flask is 
(a) ?? . ???? × ????
?? 
(b) ?? . ???? × ????
?? 
(c) ???? . ???? × ????
?? 
(d) ???? . ???? × ????
?? 
Ans : (a) ?? = 10
-3
 m
3
, N = 3.0 × 10
22
, m= 5.3 × 10
-26
 kg , ?? ?????? = 400 m /s 
?? =
1
3
????
?? ?? ?????? 2
=
1
3
×
5.3 × 10
-26
× 3.0 × 10
22
10
-3
( 400)
2
= 8.48 × 10
4
 N/m
2
. 
Q3. A flask is filled with ???????? of an ideal gas at ????
°
?? and its temperature is raised to ????
°
?? . The 
mass of the gas that has to be released to maintain the temperature of the gas in the flask at ????
°
?? 
and the pressure remaining the same is 
(a) ?? . ?? ?? 
(b) ?? . ?? ?? 
(c) ?? . ?? ?? 
(d) ?? . ?? ?? 
Ans: (d) ???? ? Mass of gas × Temperature In this problem pressure and volume remains constant so 
?? 1
?? 1
= ?? 2
?? 2
= constant 
?
?? 2
?? 1
=
?? 1
?? 2
=
( 27 + 273)
( 52 + 273)
=
300
325
=
12
13
? ?? 2
= ?? 1
×
12
13
= 13 ×
12
13
gm= 12gm 
i.e. the mass of gas released from the flask = 13gm- 12gm= 1gm . 
 
Q4. If the value of molar gas constant is ?? . ?? ?? / mole -?? , the ?? specific gas constant for hydrogen 
in ?? / mole- ?? will be 
(a) 4.15 
(b) 8.3 
(c) 16.6 
(d) None of these 
Ans : (a) Specific gas constant ?? =
 Universal gas constant ( ?? )
 Molecular weight of gas ( ?? )
=
8.3
2
= 4.15Joule /mole- K. 
Q5. At the top of a mountain a thermometer reads ?? °
?? and a barometer reads ???? ???? of ???? . At 
the bottom of the mountain these read ????
°
?? and ???? ???? of ???? respectively. Comparison of density 
of air at the top with that of bottom is 
(a) ???? /???? 
(b) ???? /???? 
(c) ???? /???? 
(d) ???? /???? 
 
Ans : (a) Ideal gas equation, in terms of density 
?? 1
?? 1
?? 1
=
?? 2
?? 2
?? 2
= constant ?
?? 1
?? 2
=
?? 1
?? 2
×
?? 2
?? 1
 
 
?
?? Top 
?? Bottom 
=
?? Top 
?? Bottom 
×
?? Bottom 
?? Top 
=
70
76
×
300
280
=
75
76
 
Q6. The equation of state of a gas is given by (?? +
?? ?? ?? ?? )?? ?? = ( ???? + ?? ) , where ?? , ?? , ?? and ?? are 
constants. The isotherms can be represented by ?? = ?? ?? ?? - ?? ?? ?? , where ?? and ?? depend only on 
temperature then 
 
(a) ?? = -?? and ?? = -?? 
(b) ?? = ?? and ?? = ?? 
(c) ?? = -?? and ?? = ?? 
(d) ?? = ?? and ?? = -?? 
Ans : (a) (?? +
?? ?? 2
?? )?? ?? = ???? + ?? ? ?? + ?? ?? 2
?? -1
= ???? ?? -?? + ?? ?? -?? ? ?? = ( ???? + ?? ) ?? -?? -
( ?? ?? 2
) ?? -1
 
By comparing this equation with given equation ?? = ?? ?? ?? - ?? ?? ?? we get ?? = -?? and ?? = -1. 
 
Page 3


Solved Examples on Kinetic Theory of 
Gases 
JEE Mains 
Single Correct 
Q1. The root mean square speed of hydrogen molecules of an ideal hydrogen gas kept in a gas 
chamber at ?? °
?? is ???????? ?? /?? . The pressure on the hydrogen gas is 
(Density of hydrogen gas is ?? . ???? × ????
-?? ???? /?? ?? , ?? atmosphere = ?? . ???? × ????
?? ?? /?? ?? ) 
(a) ?? . ?? ?????? 
(b) ?? . ?? ?????? 
(c) ?? . ?? ?????? 
(d) ?? . ?? ?????? 
Ans : (d) As ?? =
1
3
?? ?? rms
2
=
1
3
( 8.99 × 10
-2
)× ( 3180 )
2
= 3.03 × 10
5
 N/m
2
= 3.0 atm 
Q2. A flask contains ????
-?? ?? ?? gas. At a temperature, the number of molecules of oxygen are 
?? . ?? × ????
????
. The mass of an oxygen molecule is ?? . ?? × ????
-????
 ???? and at that temperature the ?????? 
velocity of molecules is ?????? ?? /?? . The pressure in ?? /?? ?? of the gas in the flask is 
(a) ?? . ???? × ????
?? 
(b) ?? . ???? × ????
?? 
(c) ???? . ???? × ????
?? 
(d) ???? . ???? × ????
?? 
Ans : (a) ?? = 10
-3
 m
3
, N = 3.0 × 10
22
, m= 5.3 × 10
-26
 kg , ?? ?????? = 400 m /s 
?? =
1
3
????
?? ?? ?????? 2
=
1
3
×
5.3 × 10
-26
× 3.0 × 10
22
10
-3
( 400)
2
= 8.48 × 10
4
 N/m
2
. 
Q3. A flask is filled with ???????? of an ideal gas at ????
°
?? and its temperature is raised to ????
°
?? . The 
mass of the gas that has to be released to maintain the temperature of the gas in the flask at ????
°
?? 
and the pressure remaining the same is 
(a) ?? . ?? ?? 
(b) ?? . ?? ?? 
(c) ?? . ?? ?? 
(d) ?? . ?? ?? 
Ans: (d) ???? ? Mass of gas × Temperature In this problem pressure and volume remains constant so 
?? 1
?? 1
= ?? 2
?? 2
= constant 
?
?? 2
?? 1
=
?? 1
?? 2
=
( 27 + 273)
( 52 + 273)
=
300
325
=
12
13
? ?? 2
= ?? 1
×
12
13
= 13 ×
12
13
gm= 12gm 
i.e. the mass of gas released from the flask = 13gm- 12gm= 1gm . 
 
Q4. If the value of molar gas constant is ?? . ?? ?? / mole -?? , the ?? specific gas constant for hydrogen 
in ?? / mole- ?? will be 
(a) 4.15 
(b) 8.3 
(c) 16.6 
(d) None of these 
Ans : (a) Specific gas constant ?? =
 Universal gas constant ( ?? )
 Molecular weight of gas ( ?? )
=
8.3
2
= 4.15Joule /mole- K. 
Q5. At the top of a mountain a thermometer reads ?? °
?? and a barometer reads ???? ???? of ???? . At 
the bottom of the mountain these read ????
°
?? and ???? ???? of ???? respectively. Comparison of density 
of air at the top with that of bottom is 
(a) ???? /???? 
(b) ???? /???? 
(c) ???? /???? 
(d) ???? /???? 
 
Ans : (a) Ideal gas equation, in terms of density 
?? 1
?? 1
?? 1
=
?? 2
?? 2
?? 2
= constant ?
?? 1
?? 2
=
?? 1
?? 2
×
?? 2
?? 1
 
 
?
?? Top 
?? Bottom 
=
?? Top 
?? Bottom 
×
?? Bottom 
?? Top 
=
70
76
×
300
280
=
75
76
 
Q6. The equation of state of a gas is given by (?? +
?? ?? ?? ?? )?? ?? = ( ???? + ?? ) , where ?? , ?? , ?? and ?? are 
constants. The isotherms can be represented by ?? = ?? ?? ?? - ?? ?? ?? , where ?? and ?? depend only on 
temperature then 
 
(a) ?? = -?? and ?? = -?? 
(b) ?? = ?? and ?? = ?? 
(c) ?? = -?? and ?? = ?? 
(d) ?? = ?? and ?? = -?? 
Ans : (a) (?? +
?? ?? 2
?? )?? ?? = ???? + ?? ? ?? + ?? ?? 2
?? -1
= ???? ?? -?? + ?? ?? -?? ? ?? = ( ???? + ?? ) ?? -?? -
( ?? ?? 2
) ?? -1
 
By comparing this equation with given equation ?? = ?? ?? ?? - ?? ?? ?? we get ?? = -?? and ?? = -1. 
 
Q7. The conversion of ideal gas into solids is 
(a) Possible only at low pressure 
(b) Possible only at low temperature 
(c) Possible only at low volume 
(d) Impossible 
Ans: (d) Because there is zero attraction between the molecules of ideal gas. 
Q8. At what temperature is the root mean square velocity of gaseous hydrogen molecules is equal 
to that of oxygen molecules at ????
°
?? 
(a) ???? ?? 
(b) ???? ?? 
(c) -???? ?? 
(d) ?? ?? 
Ans : (a) For oxygen ?? O
2
= v
3?? ?? O
2
?? O
2
 and   For hydrogen ?? H
2
= v3?? ?? H
2
?? H
2
 
 According to problem = v
3?? ?? ?? 2
?? ?? 2
= v3?? ?? ?? 2
?? ?? 2
 ?
?? ?? 2
?? ?? 2
=
?? ?? 2
?? ?? 2
?
47 + 273
32
=
?? ?? 2
2
? ?? ?? 2
=
320
32
× 2 = 20 K.
 
Q9. The root mean square speed of the molecules of a diatomic gas is ?? . When the temperature is 
doubled, the molecules dissociate into two atoms. The new root mean square speed of the atom is 
[Roorkee 1996] 
(a) v ?? ?? 
(b) ?? 
(c) ?? ?? 
(d) ?? ?? 
Ans : (c) ?? ?????? = v
3????
?? . According to problem ?? will becomes ?? /2 and ?? will becomes ?? /2 so the 
value of ?? ?????? will increase by v 4 = 2 times i.e. new root mean square velocity will be 2?? . 
 
 
 
 
 
 
 
Page 4


Solved Examples on Kinetic Theory of 
Gases 
JEE Mains 
Single Correct 
Q1. The root mean square speed of hydrogen molecules of an ideal hydrogen gas kept in a gas 
chamber at ?? °
?? is ???????? ?? /?? . The pressure on the hydrogen gas is 
(Density of hydrogen gas is ?? . ???? × ????
-?? ???? /?? ?? , ?? atmosphere = ?? . ???? × ????
?? ?? /?? ?? ) 
(a) ?? . ?? ?????? 
(b) ?? . ?? ?????? 
(c) ?? . ?? ?????? 
(d) ?? . ?? ?????? 
Ans : (d) As ?? =
1
3
?? ?? rms
2
=
1
3
( 8.99 × 10
-2
)× ( 3180 )
2
= 3.03 × 10
5
 N/m
2
= 3.0 atm 
Q2. A flask contains ????
-?? ?? ?? gas. At a temperature, the number of molecules of oxygen are 
?? . ?? × ????
????
. The mass of an oxygen molecule is ?? . ?? × ????
-????
 ???? and at that temperature the ?????? 
velocity of molecules is ?????? ?? /?? . The pressure in ?? /?? ?? of the gas in the flask is 
(a) ?? . ???? × ????
?? 
(b) ?? . ???? × ????
?? 
(c) ???? . ???? × ????
?? 
(d) ???? . ???? × ????
?? 
Ans : (a) ?? = 10
-3
 m
3
, N = 3.0 × 10
22
, m= 5.3 × 10
-26
 kg , ?? ?????? = 400 m /s 
?? =
1
3
????
?? ?? ?????? 2
=
1
3
×
5.3 × 10
-26
× 3.0 × 10
22
10
-3
( 400)
2
= 8.48 × 10
4
 N/m
2
. 
Q3. A flask is filled with ???????? of an ideal gas at ????
°
?? and its temperature is raised to ????
°
?? . The 
mass of the gas that has to be released to maintain the temperature of the gas in the flask at ????
°
?? 
and the pressure remaining the same is 
(a) ?? . ?? ?? 
(b) ?? . ?? ?? 
(c) ?? . ?? ?? 
(d) ?? . ?? ?? 
Ans: (d) ???? ? Mass of gas × Temperature In this problem pressure and volume remains constant so 
?? 1
?? 1
= ?? 2
?? 2
= constant 
?
?? 2
?? 1
=
?? 1
?? 2
=
( 27 + 273)
( 52 + 273)
=
300
325
=
12
13
? ?? 2
= ?? 1
×
12
13
= 13 ×
12
13
gm= 12gm 
i.e. the mass of gas released from the flask = 13gm- 12gm= 1gm . 
 
Q4. If the value of molar gas constant is ?? . ?? ?? / mole -?? , the ?? specific gas constant for hydrogen 
in ?? / mole- ?? will be 
(a) 4.15 
(b) 8.3 
(c) 16.6 
(d) None of these 
Ans : (a) Specific gas constant ?? =
 Universal gas constant ( ?? )
 Molecular weight of gas ( ?? )
=
8.3
2
= 4.15Joule /mole- K. 
Q5. At the top of a mountain a thermometer reads ?? °
?? and a barometer reads ???? ???? of ???? . At 
the bottom of the mountain these read ????
°
?? and ???? ???? of ???? respectively. Comparison of density 
of air at the top with that of bottom is 
(a) ???? /???? 
(b) ???? /???? 
(c) ???? /???? 
(d) ???? /???? 
 
Ans : (a) Ideal gas equation, in terms of density 
?? 1
?? 1
?? 1
=
?? 2
?? 2
?? 2
= constant ?
?? 1
?? 2
=
?? 1
?? 2
×
?? 2
?? 1
 
 
?
?? Top 
?? Bottom 
=
?? Top 
?? Bottom 
×
?? Bottom 
?? Top 
=
70
76
×
300
280
=
75
76
 
Q6. The equation of state of a gas is given by (?? +
?? ?? ?? ?? )?? ?? = ( ???? + ?? ) , where ?? , ?? , ?? and ?? are 
constants. The isotherms can be represented by ?? = ?? ?? ?? - ?? ?? ?? , where ?? and ?? depend only on 
temperature then 
 
(a) ?? = -?? and ?? = -?? 
(b) ?? = ?? and ?? = ?? 
(c) ?? = -?? and ?? = ?? 
(d) ?? = ?? and ?? = -?? 
Ans : (a) (?? +
?? ?? 2
?? )?? ?? = ???? + ?? ? ?? + ?? ?? 2
?? -1
= ???? ?? -?? + ?? ?? -?? ? ?? = ( ???? + ?? ) ?? -?? -
( ?? ?? 2
) ?? -1
 
By comparing this equation with given equation ?? = ?? ?? ?? - ?? ?? ?? we get ?? = -?? and ?? = -1. 
 
Q7. The conversion of ideal gas into solids is 
(a) Possible only at low pressure 
(b) Possible only at low temperature 
(c) Possible only at low volume 
(d) Impossible 
Ans: (d) Because there is zero attraction between the molecules of ideal gas. 
Q8. At what temperature is the root mean square velocity of gaseous hydrogen molecules is equal 
to that of oxygen molecules at ????
°
?? 
(a) ???? ?? 
(b) ???? ?? 
(c) -???? ?? 
(d) ?? ?? 
Ans : (a) For oxygen ?? O
2
= v
3?? ?? O
2
?? O
2
 and   For hydrogen ?? H
2
= v3?? ?? H
2
?? H
2
 
 According to problem = v
3?? ?? ?? 2
?? ?? 2
= v3?? ?? ?? 2
?? ?? 2
 ?
?? ?? 2
?? ?? 2
=
?? ?? 2
?? ?? 2
?
47 + 273
32
=
?? ?? 2
2
? ?? ?? 2
=
320
32
× 2 = 20 K.
 
Q9. The root mean square speed of the molecules of a diatomic gas is ?? . When the temperature is 
doubled, the molecules dissociate into two atoms. The new root mean square speed of the atom is 
[Roorkee 1996] 
(a) v ?? ?? 
(b) ?? 
(c) ?? ?? 
(d) ?? ?? 
Ans : (c) ?? ?????? = v
3????
?? . According to problem ?? will becomes ?? /2 and ?? will becomes ?? /2 so the 
value of ?? ?????? will increase by v 4 = 2 times i.e. new root mean square velocity will be 2?? . 
 
 
 
 
 
 
 
 
 
 
 
 
Q10. An ideal gas ( ?? = ?? . ?? ) is expanded adiabatically. How many times has the gas to be 
expanded to reduce the root mean square velocity of molecules 2 times 
(a) 4 times 
(b) 16 times 
(c) 8 times 
(d) 2 times 
Ans: (b) To reduce the rms velocity two times, temperature should be reduced by four times (As 
?? ?????? ? v ?? ) 
? ?? 1
= ?? ?? 2
=
?? 4
, ?? 1
= ?? 
From adiabatic law ?? ?? ?? -1
= constant we get (
?? 2
?? 1
)
?? -1
=
?? 1
?? 2
= 4 ?
?? 2
?? 1
= ( 4)
1
?? -1
 [?? = 
3/2 given] 
? ?? 2
= ?? 1
( 4)
1
3/2-1
= ?? 1
( 4)
2
= 16?? 1
 ?
?? 2
?? 1
= 16 
 
Q11. The kinetic energy of one gram mole of a gas at normal temperature and pressure is ( ?? = 
?? . ???? ?? /???????? - ?? ) 
(a) ?? . ???? × ????
?? ?? 
(b) ?? . ?? × ????
?? ?? 
(c) ?? . ?? × ????
?? ?? 
(d) ?? . ?? × ????
?? ?? 
Ans : (d) ?? =
3
2
???? =
3
2
× 8.31 × 273 = 3.4 × 10
3
 Joule 
Q12. The graph which represent the variation of mean kinetic energy of molecules with temperature 
?? °
C is 
 
Page 5


Solved Examples on Kinetic Theory of 
Gases 
JEE Mains 
Single Correct 
Q1. The root mean square speed of hydrogen molecules of an ideal hydrogen gas kept in a gas 
chamber at ?? °
?? is ???????? ?? /?? . The pressure on the hydrogen gas is 
(Density of hydrogen gas is ?? . ???? × ????
-?? ???? /?? ?? , ?? atmosphere = ?? . ???? × ????
?? ?? /?? ?? ) 
(a) ?? . ?? ?????? 
(b) ?? . ?? ?????? 
(c) ?? . ?? ?????? 
(d) ?? . ?? ?????? 
Ans : (d) As ?? =
1
3
?? ?? rms
2
=
1
3
( 8.99 × 10
-2
)× ( 3180 )
2
= 3.03 × 10
5
 N/m
2
= 3.0 atm 
Q2. A flask contains ????
-?? ?? ?? gas. At a temperature, the number of molecules of oxygen are 
?? . ?? × ????
????
. The mass of an oxygen molecule is ?? . ?? × ????
-????
 ???? and at that temperature the ?????? 
velocity of molecules is ?????? ?? /?? . The pressure in ?? /?? ?? of the gas in the flask is 
(a) ?? . ???? × ????
?? 
(b) ?? . ???? × ????
?? 
(c) ???? . ???? × ????
?? 
(d) ???? . ???? × ????
?? 
Ans : (a) ?? = 10
-3
 m
3
, N = 3.0 × 10
22
, m= 5.3 × 10
-26
 kg , ?? ?????? = 400 m /s 
?? =
1
3
????
?? ?? ?????? 2
=
1
3
×
5.3 × 10
-26
× 3.0 × 10
22
10
-3
( 400)
2
= 8.48 × 10
4
 N/m
2
. 
Q3. A flask is filled with ???????? of an ideal gas at ????
°
?? and its temperature is raised to ????
°
?? . The 
mass of the gas that has to be released to maintain the temperature of the gas in the flask at ????
°
?? 
and the pressure remaining the same is 
(a) ?? . ?? ?? 
(b) ?? . ?? ?? 
(c) ?? . ?? ?? 
(d) ?? . ?? ?? 
Ans: (d) ???? ? Mass of gas × Temperature In this problem pressure and volume remains constant so 
?? 1
?? 1
= ?? 2
?? 2
= constant 
?
?? 2
?? 1
=
?? 1
?? 2
=
( 27 + 273)
( 52 + 273)
=
300
325
=
12
13
? ?? 2
= ?? 1
×
12
13
= 13 ×
12
13
gm= 12gm 
i.e. the mass of gas released from the flask = 13gm- 12gm= 1gm . 
 
Q4. If the value of molar gas constant is ?? . ?? ?? / mole -?? , the ?? specific gas constant for hydrogen 
in ?? / mole- ?? will be 
(a) 4.15 
(b) 8.3 
(c) 16.6 
(d) None of these 
Ans : (a) Specific gas constant ?? =
 Universal gas constant ( ?? )
 Molecular weight of gas ( ?? )
=
8.3
2
= 4.15Joule /mole- K. 
Q5. At the top of a mountain a thermometer reads ?? °
?? and a barometer reads ???? ???? of ???? . At 
the bottom of the mountain these read ????
°
?? and ???? ???? of ???? respectively. Comparison of density 
of air at the top with that of bottom is 
(a) ???? /???? 
(b) ???? /???? 
(c) ???? /???? 
(d) ???? /???? 
 
Ans : (a) Ideal gas equation, in terms of density 
?? 1
?? 1
?? 1
=
?? 2
?? 2
?? 2
= constant ?
?? 1
?? 2
=
?? 1
?? 2
×
?? 2
?? 1
 
 
?
?? Top 
?? Bottom 
=
?? Top 
?? Bottom 
×
?? Bottom 
?? Top 
=
70
76
×
300
280
=
75
76
 
Q6. The equation of state of a gas is given by (?? +
?? ?? ?? ?? )?? ?? = ( ???? + ?? ) , where ?? , ?? , ?? and ?? are 
constants. The isotherms can be represented by ?? = ?? ?? ?? - ?? ?? ?? , where ?? and ?? depend only on 
temperature then 
 
(a) ?? = -?? and ?? = -?? 
(b) ?? = ?? and ?? = ?? 
(c) ?? = -?? and ?? = ?? 
(d) ?? = ?? and ?? = -?? 
Ans : (a) (?? +
?? ?? 2
?? )?? ?? = ???? + ?? ? ?? + ?? ?? 2
?? -1
= ???? ?? -?? + ?? ?? -?? ? ?? = ( ???? + ?? ) ?? -?? -
( ?? ?? 2
) ?? -1
 
By comparing this equation with given equation ?? = ?? ?? ?? - ?? ?? ?? we get ?? = -?? and ?? = -1. 
 
Q7. The conversion of ideal gas into solids is 
(a) Possible only at low pressure 
(b) Possible only at low temperature 
(c) Possible only at low volume 
(d) Impossible 
Ans: (d) Because there is zero attraction between the molecules of ideal gas. 
Q8. At what temperature is the root mean square velocity of gaseous hydrogen molecules is equal 
to that of oxygen molecules at ????
°
?? 
(a) ???? ?? 
(b) ???? ?? 
(c) -???? ?? 
(d) ?? ?? 
Ans : (a) For oxygen ?? O
2
= v
3?? ?? O
2
?? O
2
 and   For hydrogen ?? H
2
= v3?? ?? H
2
?? H
2
 
 According to problem = v
3?? ?? ?? 2
?? ?? 2
= v3?? ?? ?? 2
?? ?? 2
 ?
?? ?? 2
?? ?? 2
=
?? ?? 2
?? ?? 2
?
47 + 273
32
=
?? ?? 2
2
? ?? ?? 2
=
320
32
× 2 = 20 K.
 
Q9. The root mean square speed of the molecules of a diatomic gas is ?? . When the temperature is 
doubled, the molecules dissociate into two atoms. The new root mean square speed of the atom is 
[Roorkee 1996] 
(a) v ?? ?? 
(b) ?? 
(c) ?? ?? 
(d) ?? ?? 
Ans : (c) ?? ?????? = v
3????
?? . According to problem ?? will becomes ?? /2 and ?? will becomes ?? /2 so the 
value of ?? ?????? will increase by v 4 = 2 times i.e. new root mean square velocity will be 2?? . 
 
 
 
 
 
 
 
 
 
 
 
 
Q10. An ideal gas ( ?? = ?? . ?? ) is expanded adiabatically. How many times has the gas to be 
expanded to reduce the root mean square velocity of molecules 2 times 
(a) 4 times 
(b) 16 times 
(c) 8 times 
(d) 2 times 
Ans: (b) To reduce the rms velocity two times, temperature should be reduced by four times (As 
?? ?????? ? v ?? ) 
? ?? 1
= ?? ?? 2
=
?? 4
, ?? 1
= ?? 
From adiabatic law ?? ?? ?? -1
= constant we get (
?? 2
?? 1
)
?? -1
=
?? 1
?? 2
= 4 ?
?? 2
?? 1
= ( 4)
1
?? -1
 [?? = 
3/2 given] 
? ?? 2
= ?? 1
( 4)
1
3/2-1
= ?? 1
( 4)
2
= 16?? 1
 ?
?? 2
?? 1
= 16 
 
Q11. The kinetic energy of one gram mole of a gas at normal temperature and pressure is ( ?? = 
?? . ???? ?? /???????? - ?? ) 
(a) ?? . ???? × ????
?? ?? 
(b) ?? . ?? × ????
?? ?? 
(c) ?? . ?? × ????
?? ?? 
(d) ?? . ?? × ????
?? ?? 
Ans : (d) ?? =
3
2
???? =
3
2
× 8.31 × 273 = 3.4 × 10
3
 Joule 
Q12. The graph which represent the variation of mean kinetic energy of molecules with temperature 
?? °
C is 
 
Ans : (c) Mean K.E. of gas molecule ?? =
3
2
???? =
3
2
?? ( ?? + 273) where ?? = temperature is in kelvin and 
?? = is in centigrade 
? ?? =
3
2
???? +
3
2
× 273?? ?? = Boltzmann's constant 
By comparing this equation with standard equation of straight line ?? = ???? + ?? 
We get ?? =
3
2
?? and ?? =
3
2
273?? . So the graph between ?? and ?? will be straight line with positive 
intercept on ?? -axis and positive slope with ?? -axis. 
 
Q13. An air bubble of volume ?? ?? is released by a fish at a depth ?? in a lake. The bubble rises to the 
surface. Assume constant temperature and standard atmospheric pressure ?? above the lake. The 
volume of the bubble just before touching the surface will be (density of water is ?? ) 
(a) ?? ?? 
(b) ?? ?? ( ?????? /?? ) 
(c) 
?? ?? (?? +
??????
?? )
 
(d) ?? ?? (?? +
?????? ?? ) 
 
Ans: (d) According to Boyle's law multiplication of pressure and volume will remains constant at the 
bottom and top. 
If ?? is the atmospheric pressure at the top of the lake and the volume of bubble is ?? then from 
?? 1
?? 1
= ?? 2
?? 2
 
 ( ?? + h???? ) ?? 0
= ???? ? ?? = (
?? + h????
?? )?? 0
? ?? = ?? 0
[1 +
???? h
?? ]
 
Q14. A gas is filled in the cylinder shown in the figure. The two pistons are joined by a string. If the 
gas is heated, the pistons will 
(a) Move towards left 
(b) Move towards right 
(c) Remain stationary 
(d) None of these 
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FAQs on Kinetic Theory of Gases Solved Examples - Physics for JEE Main & Advanced

1. What is the Kinetic Theory of Gases?
Ans. The Kinetic Theory of Gases is a model that describes the behavior of gases by considering their particles as point masses in constant, random motion.
2. How does temperature affect the kinetic energy of gas particles?
Ans. As temperature increases, the kinetic energy of gas particles also increases, leading to higher average speeds and more frequent collisions.
3. What is the relationship between pressure and volume in the Kinetic Theory of Gases?
Ans. According to the Kinetic Theory of Gases, the pressure of a gas is directly proportional to its temperature and the number of gas particles, and inversely proportional to its volume.
4. How does the Kinetic Theory of Gases explain the ideal gas law?
Ans. The Kinetic Theory of Gases provides a theoretical basis for the ideal gas law by assuming that gas particles are in constant motion and have negligible volume compared to the container they are in.
5. What are some real-world applications of the Kinetic Theory of Gases?
Ans. The Kinetic Theory of Gases is used in various fields such as meteorology, chemistry, and engineering to understand and predict the behavior of gases under different conditions.
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