Pressure Of An Ideal Gas

Let's first review the definition of an ideal gas before learning how to compute its pressure. Put simply, an ideal gas is a theoretical gas in which there is no interparticle interaction and the gas particles move arbitrarily. There is no such thing as an ideal gas. It is based on the ideal gas equation, a simplified formula that can be analysed using statistical mechanics, about which we shall learn more. Most gases are assumed to behave as ideal gases under conventional pressure and temperature conditions.

This Story also Contains

1. The pressure of an Ideal Gas
2. Instantaneous Velocity
3. What is Collision Frequency?
4. What is Change in Momentum?
5. What is Force on the Wall?
6. Pressure
7. Solved Examples Based on Pressure of an Ideal Gas
8. Summary

This article will cover the concept of the Pressure of an Ideal Gas. This concept we study in the chapter on the kinetic theory of gases, which is a crucial chapter in Class 11 physics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), National Eligibility Entrance Test (NEET), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), two questions have been asked on this concept. And for NEET one question was asked from this concept.

Let's read this entire article to gain an in-depth understanding of the Pressure of an Ideal Gas.

The pressure of an Ideal Gas

Consider an ideal gas (consisting of N molecules each of mass m) enclosed in a cubical box of side L as shown in the figure below.

Instantaneous Velocity

Any molecule of gas moves with velocity v→ in any direction

where v→=vxi^+vyj^+vzk^

Due to the random motion of the molecule

vx=vy=vz As v=vx2+vy2+vz2⇒v=3vx2=3vy2=3vz2

What is the Time During a Collision?

The time during a collision Time between two successive collisions with the wall A₁

_{I.e Δt= Distance travelled by molecule between two successive collision Velocity of molecule or Δt=2Lvx}

What is Collision Frequency?

Collision frequency (n): It means the number of collisions per second.

I.e n=1Δt=vx2L

What is Change in Momentum?

Change in momentum: This molecule collides with A₁ wall (A1) with velocity v_x and rebounds with velocity (-v_x)
The change in momentum of the molecule is given by

Δp=(−mvx)−(mvx)=−2mvx

As the momentum remains conserved in a collision,

Δpsystem =0Δpsystem =Δpmolecule +Δpwall =0Δpwall =−Δpmolecule
the change in momentum of wall A1 is Δp=2mvx

the change in momentum of wall A₁ is Δp=2mvx

What is Force on the Wall?

Force on the wall: Force exerted by a single molecule on the A₁ wall is equal to the rate at which the momentum is transferred to the wall by this molecule.

i.e. FSingle molecule =ΔpΔt=2mvx(2L/vx)=mvx2L

The total force on the wall A₁ due to N molecules

Fx=mL∑vx2=mL(vx12+vx22+vx32+…)=mNLvx2―

where vx2―= mean square of x component of the velocity.

Pressure

Pressure- As pressure is defined as force per unit area, hence the pressure on A₁ wall

Px=FxA=mNALvx2―=mNVvx2― As vx2―=vy2―=vz2― So v2―=vx2―+vy2―+vz2v2⇒vx2―=vy2―=vz2―=v33

So, the Total pressure inside the container is given by

P=13mNVv2―=13mNVvrms2 (where vrms=v2― )

Using total mass= M = mN

Pressure due to an ideal gas is given as

P=13ρvrms2=13(MV)⋅vrms2

where

m= mass of one molecule
N= Number of the molecule
vrms2=v12+v22+……….n
vrms= RMS velocity

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Solved Examples Based on Pressure of an Ideal Gas

Example 1: Based on the kinetic theory of gases, the gas exerts pressure because of its molecules:

1) Continuously lose their energy till it reaches the wall.

2) Suffer change in momentum when impinge on the walls of the container.

3) are attracted by the walls of the container.

4) Continuously stick to the walls of the container.

Solution:

Based on the kinetic theory of gases, molecules suffer a change in momentum when impinging on the walls of the container. Due to this, they exert a force resulting in exerting pressure on the walls of the container.

Hence, the answer is the option (2).

Example 2: When a gas filled in a closed vessel is heated by raising the temperature 1∘C, its pressure increases 0.4%. The initial temperature of the gas is____ K.

1) 260

2) 250

3) 300

4) 350

Solution:

For a closed vessel,

V= constant ∴PαTΔPP=ΔTT0.4100=1 T T=250 K

Hence, the answer is (2).

Example 3: In an ideal gas at temperature T, the average force that a molecule applies on the walls of a closed container depends on T as T^q. A good estimate for q is :

1) 1

2) 2

3) 0.5

4) 0.25

Solution:

Force ∝dpdt (Rate of change of momentum)
=mv2L
L is the length of the container and v2 is the mean square velocity
v2αT∴FαT

Hence, the Value of q =1

Hence, the answer is option (1).

Example 4: Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as V^q, where V is the volume of the gas. The value of q is :

(γ=CpCv)
1) 3γ+56
2) 3γ−56
3) γ+12
4) γ−12

Solution:

Average time of collision : τ=1nπ2vrmsd2 since vrms∝T and n∝1V
For an adiabatic process :
TVγ−1= constant T∝V(1−γ2)
∴τ∝VT∴τ∝V(1+γ2)

Hence, the answer is the option (3).

Example 5: An ideal gas at atmospheric pressure is adiabatically compressed so that its density becomes 32 times its initial value. If the final pressure of a gas is 128 atmospheres, the value of ' γ ' of the gas is :

1) $1.5$

2) $1.4$

3) $1.3$

4) $1.6$

Solution:

Volume of the gas, v= mass density =mρ and using (PV)γ= constant P′P=VVγ=(mρmρ)γ=(ρ′ρ)γ

Or 128=(32)7
∴γ=log32128∴γ=75=1.4

Hence, the answer is option (2).

Summary

The fundamental part of thermodynamics is understanding that ideal gases have pressure - this is the key to how they bounce back from their container walls. To explain how much pressure is being applied to the walls of the container temperature, volume and number of molecules are of significance. Such relations between them will make us know why gases behave differently in different situations and important fields like engineering and meteorology cannot continue without understanding these facts in mastering the principle of ideal.