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Solved Examples: Kinetic Theory of Gases | Mock Tests for JEE Main and Advanced 2025 PDF Download

PROBLEM 1 (JEE MAIN):-
Calculate the number of molecules in 2 x 10-6 m3 of a perfect gas at 27°C and at a pressure of 0.01 m of mercury. Mean K.E of a molecule at 27°C = 4 x 10-11 J and g = 9.8 ms-2.
Solution:-
P = (1/3) (M/V) Cor PV = (1/3) MC2
But M = (m) (n)
Where ‘m’ is mass of one molecule and n is the number of molecules.
PV = (1/3) (m) C2
n = 3PV/mC2 = [3/2 PV] / [(1/2) mC2]
Here, P = 0.01 m of mercury = 0.01 x 13600 x 9.8Nm-2
V =  2 x 10-6 m3 = 1332.8 Nm-2
½ mc2 = 4 x 10-11J
Solved Examples: Kinetic Theory of Gases | Mock Tests for JEE Main and Advanced 2025
Solved Examples: Kinetic Theory of Gases | Mock Tests for JEE Main and Advanced 2025
Thus from the above observation we conclude that, the number of molecules would be 9.996 x 107.

Problem 2:-
You are throwing a birthday party and decide to fill the room with helium balloons. You also want to have a few larger balloons to put at the door. The smaller balloons are filled occupy 0.240 m3 when the pressure inside them is 0.038 atm and the temperature of the room is 70° F. What pressure should you fill the larger balloons to so that they occupy 0.400 m3?
Solution:-
The temperature of the room is assumed to be held constant, so it is extraneous information. Since you are dealing with volume and pressure, you would use Boyle's Law.
In accordance to data,
P1= 0.038 atm, V1= 0.240 m3
V2=0.400 m3
We have to find out V2.
Substitute the vale of P1, V1 and V2 in equation P1V1=P2V2
P1V1=P2V2
(0.038 atm)(0.240 m3) = P2(0.400 m3)
0.00912 = 0.400 P2             (units cancel out so that pressure will be in atm)
P2 = 0.0228 atm
From the above observation we conclude that, the pressure would be 0.0228 atm.

Problem 3:-
Find the RMS speed of a sample of neon gas at 80° F.
Solution:-
First convert Fahrenheit to Celsius.
°C = (°F- 32)/1.8
= (80 -32)/1.8
= 48/1.8
= 26.6
Convert Celsius to Kelvin.
K = °C + 273.15  
= 26.6 + 273.15
= 299.75 K
Substitute the known information into the equation for RMS speed and solv, we get,.
vrms = √3RT/M
= √3(8.3144) (299.75)/ (20.179)
= 19.2 m/s
From the above observation we conclude that, the RMS speed of a sample of neon gas at 80° F would be 19.2 m/s.

The document Solved Examples: Kinetic Theory of Gases | Mock Tests for JEE Main and Advanced 2025 is a part of the JEE Course Mock Tests for JEE Main and Advanced 2025.
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FAQs on Solved Examples: Kinetic Theory of Gases - Mock Tests for JEE Main and Advanced 2025

1. What is the kinetic theory of gases?
Ans. The kinetic theory of gases is a scientific model that explains the behavior of gases based on the motion of their particles. It states that gases are composed of small particles in constant random motion, and their kinetic energy determines their temperature and pressure.
2. How does the kinetic theory of gases explain the pressure exerted by gases?
Ans. According to the kinetic theory of gases, gas particles are in constant motion and collide with the walls of the container. These collisions create a force on the walls, resulting in pressure. The more frequent and intense the collisions, the higher the pressure exerted by the gas.
3. What is the relationship between temperature and the kinetic energy of gas particles?
Ans. The kinetic theory of gases states that the temperature of a gas is directly proportional to the average kinetic energy of its particles. As the temperature increases, the particles move faster, and their kinetic energy increases.
4. How does the kinetic theory of gases explain the expansion and contraction of gases with changes in temperature?
Ans. According to the kinetic theory of gases, when the temperature of a gas increases, the particles move faster and spread out, causing the gas to expand. Conversely, when the temperature decreases, the particles slow down, and the gas contracts or occupies less volume.
5. Can the kinetic theory of gases explain deviations from ideal gas behavior?
Ans. Yes, the kinetic theory of gases can explain deviations from ideal gas behavior. It considers ideal gases as having perfectly elastic collisions and negligible volume for the gas particles. Deviations from ideal behavior occur when the gas particles interact with each other or experience attractive or repulsive forces, which affect their motion and behavior.

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