At a given temperature the volume occupied by a fixed mass of a gas is inversely proportional to its pressure.
\(V\alpha \frac { 1 }{ P }\) at constant T& n.
Mathematical form: P₁V₁ = P₂ V₂ = K

Gases which obey Boyle's Law and Charle's law or ideal gas equation PV = nRT are called ideal gases.
All gases whose behaviour is consistent with the assumption of kinetic theory of gases under all condition are called ideal gases.

a) Hydrogen is the lightest gas in the atmosphere. So it rises up and other gases which are heavier like O₂ & N₂ come down towards the surface of the earth according to Graham's law of diffusions \(\mathrm{r} \alpha \sqrt{\frac{1}{M}}\). Hydrogen diffuses very fast. Its mean speed in greater than the escape velocity from the earth. As a consequence, H₂ would have escaped from the atmosphere long time ago.
b) The acceleration due to gravity 'g' on moon surface is small. The value of escape velocity is also small. The molecules of the atmospheric gases on the moon's surface have thermal velocities greater than the escape velocity.
All the molecules have escaped. So the atmosphere is so thin.

The passenger in a moving bus falls in the forward direction, when brakes are applied suddenly is in accordance with Newton's first law of motion.
When breaks are applied, the automobile comes to rest but the passenger due to inertia of motion tend to continue to move in forward direction hence fall forward.
The movement of helium balloon in the opposite direction is due to the difference in the density of surrounding air. When the car stops suddenly, the air moves forward due to inertia of motion.
As a result the air in the front of the car is more dense that at the rear of the car creating more pressure on the front side of the balloon. This causes the balloon move towards the rear of the car where air is less dense.
The body is at rest but the car is forward motion, due to inertia the passengers are pushed toward the front of the car on forward acceleration.

T₁ = 15^oC + 273 T₂= 38 + 273
T₁ = 288 K  T₂ = 311 K
V₁ = 2.58 dm³ V₂ = ?
(P = 1 atm constant)
\(\frac { { V }_{ 1 } }{ { T }_{ 1 } } =\frac { { V }_{ 2 } }{ { T }_{ 2 } } \)
\({ V }_{ 2 }=\left( \frac { { V }_{ 1 } }{ { T }_{ 1 } } \right) \times { T }_{ 2 }\)

V₂ = 2.78 dm³ i.e. volume increased from 2.58 dm³ to 2.78 dm³.