In view of the coronavirus pandemic, we are making LIVE CLASSES and VIDEO CLASSES completely FREE to prevent interruption in studies
Chemistry > States of Matter > Ideal Gas Equations
States of Matter

Ideal Gas Equations

We saw how the gas laws depict the behaviour of a gas under certain conditions of temperature and pressure. can we somehow combine these laws into one equation? The answer is yes and the equation that determines the behaviour of an ideal gas is the Ideal gas equation. Let’s learn more!

Suggested Videos

Play
Play
Play
Arrow
Arrow
ArrowArrow
Solid State
Gaseous State and its Characteristics
Postulates and Problems of Kinetic Theory of Gases Hin
Slider

 

Ideal Gas Equation

The ideal gas equation is a combination of all the individual gas laws. Boyle’s Law, Charle’s law, Gay-Lussac’s Law and Avogadro’s Laws are the basis of the Ideal gas equation.

After the analysis of the experiments in various gas laws, we can understand the idea behind the ideal behaviour of gases. The ideal gases perfectly obey the Ideal Gas Laws. The law of Ideal gases states that the volume of a specified amount of gas is inversely proportional to pressure and directly proportional to volume and temperature. Before deriving the Ideal Gas Law lets revise what various Gas Laws say.

Browse more Topics under States Of Matter

Boyle’s Law

It states that at a constant temperature, the volume of a gas is indirectly proportional to pressure. This implies that with an increase in pressure the volume of the gas will decrease. Mathematically Boyle’s law is summarized as V \(\propto\) 1/P

Charle’s Law

According to the Charle’s Law, at constant pressure, the volume of a gas is directly proportional to temperature. This means that with increasing temperature, the volume of the gas increases. Mathematically Charle’s Law can be written as V \(\propto\) T

Avogadro’s Law

Avogadro’s law states that at standard temperature and pressure conditions, the volume of a gas is directly proportional to the number of molecules of the gas. Therefore, mathematically we write it as V \(\propto\) n

The Equation

Now putting all the three laws together, we find that at standard conditions, Volume of a gas can be represented as follows:

V  \(\propto\)  T × n × 1/P

V = R × T × n × (1/P)

R, here is a proportionality constant. Therefore we can write, P V = n R T or R = PV/nT. Also called as the Gas Constant, R is same for all gases. This is therefore also called Universal Gas Constant. From the analysis of the three Gas laws, we get the Ideal Gas Equation:

PV=nRT

Therefore if the number of moles (n) is 1 we get pV= RT. Now taking the value of m/M for n we write: pV=(m/M) RT ; (n=m/M)

Since pV= mRT /M = (mRT)/VM

Therefore, p = (dRT)/M   (Since, d =m/V)

Hence we have: d = (pM) /RT

M= dRT/P. This equation brings us to the conclusion that under unchanged conditions of temperature, and pressure the density of a gas is directly proportional to the molar mass of the gas.

Ideal Gas Behaviour

Many properties of Ideal gases are very closely similar to real gases. Some of these properties specific to Ideal gases are:

  • Ideal gas constitutes a large number of molecules that are identical to each other.
  • These molecules move in random motion and obey Newton’s Laws of Motion.
  • The volume occupied by the gas is larger as compared to that of the volume occupied by the molecules.

It is only during a collision that molecules experience external forces. These collisions are elastic in nature and end in very short intervals of time.

Value of R

R is the gas constant or proportionality constant in the Ideal Gas Equation. Mathematically, if you need to find the value of any variable, then you can do so if you have the other values. In the Ideal gas law equation pV = nRT, we can write R =  pV/ nT. Now at Standard Temperature Pressure conditions i.e. when T = 273.15 K and p = 1 bar the volume is 22.710 L mol-1 and number of moles(n) is 1, then R = (105 Pa)(22.71×10-3 m3) / (1 mol)(273.15K)

= 8.314 Pa m3K-1 mol-1

= 8.314 JK-1 mol-1 

Combined Gas Law

Now if the physical conditions of temperature, pressure and volume show variation then the initial values shall be T1, pand V1 while the final values are T2, pand V2 then, we take, p1V1/T= nR and p2V2/T= nR. Therefore we have,

p1V1/T1  = p2V2/T2

This equation is known as the Combined Gas Law.

Solved  Example For You

Q: At 45° C and 900mm of Hg pressure a gas occupies 500ml volume. What will be its pressure at a height where the temperature is 15º C and volume is 750mL.

Solution: Here we use the equation of Combined Gas Law, p1V1/T1  = p2V2/T2

Therefore we have: p= p1VT2/TV2

Hence, p= 900 mmHg, T= 45 +273= 318 K, V= 500 mL, T2= 15+273 = 288 and V= 750 mL

p2 = (900 mm Hg) × (500mL) × 288 / 318 K × 750 mL = 543.96 mm Hg

Therefore the height will be = 543.96 mm

Share with friends

Customize your course in 30 seconds

Which class are you in?
5th
6th
7th
8th
9th
10th
11th
12th
Get ready for all-new Live Classes!
Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes.
tutor
tutor
Ashhar Firdausi
IIT Roorkee
Biology
tutor
tutor
Dr. Nazma Shaik
VTU
Chemistry
tutor
tutor
Gaurav Tiwari
APJAKTU
Physics
Get Started

Leave a Reply

avatar
  Subscribe  
Notify of

Stuck with a

Question Mark?

Have a doubt at 3 am? Our experts are available 24x7. Connect with a tutor instantly and get your concepts cleared in less than 3 steps.
toppr Code

chance to win a

study tour
to ISRO

Download the App

Watch lectures, practise questions and take tests on the go.

Get Question Papers of Last 10 Years

Which class are you in?
No thanks.