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Test: Kinetic Theory (29 Sep) - JEE MCQ
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Test: Kinetic Theory (29 Sep) - Question 1
The P-V diagram of a diatomic gas is a straight line passing through origin. The molar heat capacity of the gas in the process will be
Detailed Solution for Test: Kinetic Theory (29 Sep) - Question 1
or 
constant Molar heat capacity in the process 
constant is 
Here 
(For diatomic gas) 
Test: Kinetic Theory (29 Sep) - Question 2
The ratio of the adiabatic to isothermal elasticities of a triatomic (non-linear) gas is
Detailed Solution for Test: Kinetic Theory (29 Sep) - Question 2
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Test: Kinetic Theory (29 Sep) - Question 3
The specific heats, Cp and CV of gas of diatomic molecules, A, is given (in units of J mol −1K−1) by 29 and 22, respectively. Another gas of diatomic molecules, B, has the corresponding values 30 and 21. If they are treated as ideal gases, then
Detailed Solution for Test: Kinetic Theory (29 Sep) - Question 3
and 
of A are 29 and 22 and 
and 
of 
are 30 and 
Test: Kinetic Theory (29 Sep) - Question 4
One mole of a gas occupies 
lit at N.T.P. Calculate the difference between two molar specific heats of the gas. 
lit at N.T.P. Calculate the difference between two molar specific heats of the gas.
Detailed Solution for Test: Kinetic Theory (29 Sep) - Question 4
litre 
Test: Kinetic Theory (29 Sep) - Question 5
If for hydrogen 
and for the nitrogen 
, where 
and 
refer to specific heat per unit mass at constant pressure and specific heat per unit mass at constant volume respectively. The relation between 
and 
is
and for the nitrogen , where and refer to specific heat per unit mass at constant pressure and specific heat per unit mass at constant volume respectively. The relation between and is
Detailed Solution for Test: Kinetic Theory (29 Sep) - Question 5
or 
Test: Kinetic Theory (29 Sep) - Question 6
If the degree of freedom of a gas molecule is 
, then the ratio of two specific heats is given by
, then the ratio of two specific heats is given by
Detailed Solution for Test: Kinetic Theory (29 Sep) - Question 6
Test: Kinetic Theory (29 Sep) - Question 7
The average translational kinetic energy of a molecule in a gas becomes equal to 0.69eV at temperature about, [Boltzmann's constant = 138 × 10−23 J K−1]
Detailed Solution for Test: Kinetic Theory (29 Sep) - Question 7
Test: Kinetic Theory (29 Sep) - Question 8
and He are enclosed in identical containers under the similar conditions of pressure and temperature. The gases will have
Detailed Solution for Test: Kinetic Theory (29 Sep) - Question 8
Here 
is constant
Test: Kinetic Theory (29 Sep) - Question 9
Let 
denote the ratio of specific heat for an ideal gas. Then choose the correct expression for number of degrees of freedom of a molecule of the same gas.
Detailed Solution for Test: Kinetic Theory (29 Sep) - Question 9
= Total Energy, 
= Degree of freedom of molecule, We Know,
Molar heat capacity at constant volume 
= Molar heat capacity at constant pressure 
= Ratio of specific heats 
Test: Kinetic Theory (29 Sep) - Question 10
Two identical calorimeters 
and 
contain an equal quantity of water at 
. A 
piece of metal 
of specific heat 
is dropped into 
and 
piece of metal 
is dropped into 
. The equilibrium temperature in 
is 
and that in 
is 
. The initial temperature of both the metals was 
. The specific heat of metal 
(in 
) is
and contain an equal quantity of water at . A piece of metal of specific heat is dropped into and piece of metal is dropped into . The equilibrium temperature in is and that in is . The initial temperature of both the metals was . The specific heat of metal (in ) is
Detailed Solution for Test: Kinetic Theory (29 Sep) - Question 10
be the specific heat of the metal 
, then
Test: Kinetic Theory (29 Sep) - Question 11
One mole of a diatomic gas undergoes a process 
, where 
and 
are constants. The translational kinetic energy of the gas when 
is given by
, where and are constants. The translational kinetic energy of the gas when is given by
Detailed Solution for Test: Kinetic Theory (29 Sep) - Question 11
Test: Kinetic Theory (29 Sep) - Question 12
Temperature at which the kinetic energy of gas molecule is half of the value of kinetic energy at 
Detailed Solution for Test: Kinetic Theory (29 Sep) - Question 12
is Boltzmann's constant.
Test: Kinetic Theory (29 Sep) - Question 13
A nitrogen molecule has some rms speed at 
on the surface of the earth. With this speed, it goes straight up. If there is no collisions with other molecules, the molecule will rise up to a height of
Detailed Solution for Test: Kinetic Theory (29 Sep) - Question 13
Detailed Solution for Test: Kinetic Theory (29 Sep) - Question 14
Test: Kinetic Theory (29 Sep) - Question 15
At what absolute temperature 
is the root mean square speed of a hydrogen molecule equal to its escape velocity from the surface of the moon? The radius of moon is 
is the acceleration due to gravity on moon's surface, 
is the mass of a hydrogen molecule and 
is the Boltzmann constant
is the root mean square speed of a hydrogen molecule equal to its escape velocity from the surface of the moon? The radius of moon is is the acceleration due to gravity on moon's surface, is the mass of a hydrogen molecule and is the Boltzmann constant
Detailed Solution for Test: Kinetic Theory (29 Sep) - Question 15
, we require
or 
, which is choice (d).|
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