Ideal Gas Equation

Ideal Gas Equation

Edited By Vishal kumar | Updated on Nov 12, 2024 01:31 AM IST

Compared to solid and liquid properties, the properties of gases are simpler to comprehend. This is primarily due to the fact that molecules in a gas are dispersed widely apart and barely interact with one another unless they collide. The molecules that make up an ideal gas are those that have no external forces at work other than the forces created when they collide with the container wall and with each other.

In this article, we will cover the concept of the Ideal gas equation. This concept is part of the chapter 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 and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of twelve questions have been asked on this concept. And for NEET ten questions were asked from this concept.

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

Ideal Gas Equation

The equation which relates the pressure (P), volume (V) and temperature (T) of the given state of an ideal gas is known as an ideal gas equation or equation of state.

From Boyle’s law, we get V1P...(1)

and From Charle’s Law, we get VT(2)

And from Avogadro’s Law, we get Vn.(3)

And from equation (1), (2), (3)

we can write

VnTP or PVnT= constant or PVnT=R( where R is proportionality constant ) or PV=nRT

So, Ideal Gas Equation is given as

P V = n R T

where

T= Temperature

P= pressure of ideal gas

V= volume

n= numbers of mole

R = universal gas constant

Universal gas constant (R)

At S.T.P. the value of the universal gas constant is the same for all gases. And its value is given as

R=8.31 Joule mole × Kelvin =2cal mole × Kelvin

And its Dimension is : [ML2T2θ1]

Boltzman's constant (k)

It is represented by the per mole gas constant.

k=RN=8.316.023×1023=1.38×1023 J/K

Specific gas constant (r)

It is represented by per gram gas constant.

i.e. r=RM

It's unit is Joule gm× kelvin

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Solved Examples Based on the Ideal gas equation

Example 1: Which of the following shows the correct relationship between the pressure ‘P’ and density ρ of an ideal gas at constant temperature?

1)

2)

3)

4)

Solution:

Ideal gas equation

P V=n R T
wherein

T = Temperature

P= pressure of ideal gas

V= volume

n= numbers of mole

R = universal gas constant

For an ideal gas

PM=ρRT

Hence, the graph between P&ρ at constant temperature is straight line passing through origin.

Hence, the answer is option (4).

Example 2: ‘ n’ moles of an ideal gas undergoes a process A→B as shown in the figure. The maximum temperature of the gas during the process will be :

1) 9P0V04nR
2) 3P0V02nR
3) 9P0V02nR
4) 9P0V0nR

Solution:

At any point between A & B, we can write a relation between P & V by using the equation of the straight-line

VV0=2V0V0P02P0(P2P0)VV0=V0P0(P2P0)P(V0P0)+2V0=VV0P=P0V0(V3V0)

From the ideal gas equation

PV=nRTnRTV=P0V0(V3V0)T=P0nRV0(V23V0V)

For temperature to be maximum at any point dTdV=0

2V3V0=0V=3V02Tmax =P0nRV0(94V0292V02)=P0nRV094V02=94P0V0nR

Hence, the answer is option option (1).

Example 3: One mole of an ideal monoatomic gas is taken along the path ABCA as shown in the PV diagram. The maximum temperature attained by the gas along the path BC is given by :

1) 2516p0v0R
2) 258p0v0R
3) 254p0v0R
4) 58p0v0R

Solution:

Ideal gas equation

PV = nRT
wherein

T= Temperature

P= pressure of ideal gas

V= volume

n= numbers of mole

R = universal gas constant

equation of the given path is

p3p0=2p0v0(vv0)p=2p0v0v+5p0PV=RT or P=RT/VT=(2p0/Rv0)v2+5p0Rv

for T to be maximum

dT/dv=02p0Rv02v+5p0/R=0v=5v0/4Tmax=25p0v08R

Hence, the answer is option (2)

Example 4: This is true for an ideal gas.

(1) Molecule of gas is identical spherical rigid and perfectly elastic point mass.

(2) There is always some attractive and repulsive force acting between gas molecules.

(3) The density of a gas is constant at all points of the container molecules

1) only 1

2) 1 and 3

3) 2 and 3

4) only 3

Solution:

For ideal gas

  1. The molecules of a gas are identical, spherical, rigid and perfectly elastic point masses (It means that when they collide with each other, then there is no loss of energy during collision).
  2. The density of gas does not change at any point in the container.
  3. No attractive or repulsive force acts between gas molecules.

Hence, the answer is the option 2

Example 5: The temperature of an open room of volume 30 m3 increases from 17C to 27C due to the sunshine. The atmospheric pressure in the room remains 1×105 Pa. If ni and nf are the number of molecules in the room before and after heating, then nfni will be :

1) −1.61×1023

2) 1.38×1023

3) 2.5×1025

4) −2.5×1025

Solution:

PV=nRTni=PVRTi,nf=PVRTfnfni=PVR(1Tf1Ti)=105×308.31×(13001290)nfni=105×308.31×10300×290=105290×8.31
change in the Number of molecules
=105×6.023×1023290×8.31=2.5×1025

Hence, the answer is the option (4).

Summary

Not all gases can be represented by the ideal gas equation; only ideal gases can. It cannot be used for gases that are exceptionally large, exhibit strong intermolecular forces, collide in an inelastic manner, or lack kinetic energy proportional to the ideal gas's absolute temperature. Thus, there are restrictions on the Ideal Gas Equation. Because the molecular mass and the intermolecular forces are important under these circumstances, it is ineffective at low temperatures, high densities, and extremely high pressures. Heavy gases, such as refrigerants, and gases with extremely high intermolecular forces, such as water vapour, cannot be treated with it.

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