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Important Gaseous State Formula For JEE

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Last updated date: 25th Apr 2024
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Gaseous State

The gaseous state is an integral part of physical chemistry. This gaseous state provides an insight into the specifications of gases and the formulae related to it, in terms of physical chemistry. This gaseous state IIT JEE notes provides a student with better clarity into the essential parts which are to be prepared for cracking the exams. 


Specification of Gaseous State

Mass (m): 

The mass of the gas is identified with the number of moles as 

n = w/M 

Where n = number of moles 

w = mass of gas in grams 

M = sub-atomic mass of the gas 


Volume (V): 

Gases consume the whole space accessible to them. And hence, the gas volume implies the volume of the compartment wherein the gas is encased. 

Units of Volume: Volume is commonly communicated in liter (L), cm³ and dm³

1m³ = 1000 litre = 10³ dm³ = (10)6 cm³. 


Pressure: 

The weight of the gas is because of its crashes with dividers of its compartment. For example, the power applied by the gas per unit region on the dividers of the holder is equivalent to its weight. 

Pressure = Force/Area = (Mass * Acceleration) / Area

Weight is applied by gas because of the dynamic vitality of its particles. 

As temperature builds, the motor vitality of atoms expands, which brings about an increment in the weight of the gas. Along these lines, the weight of any gas is straightforwardly relative to its temperature. The weight of a gas is communicated in 

atm, Pa, Nm(–2), bar, and lb/In2 (psi). 

760 mm = 1 atm = 10132.5 KP­a = 101325 Pa = 101325 Nm(–2) 

760 mm of Hg = 1.01325 bar = 1013.25 millibar = 14.7 lb/2n2 (psi) 


Temperature (T): 

Temperature is a physical quantity that represents coldness and hotness. The SI unit of temperature is Kelvin. °C and °F are the two different units utilized for estimating temperature. On the Celsius scale water freezes at 0°C and boils at 100°C whereas in the Kelvin scale water freezes at 273 K and bubbles at 373 K. 

K = °C + 273.5 

F = (9/5) °C + 32

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An image from gaseous state iit jee notes.


Boyle's Law

At constant temperature, the weight of a fixed amount, specifically the number of moles n of gas differs conversely with its volume. 

Vα1/P 

Vα1P

P1V1 = P2V2 


Charles' Law

At constant pressure, the volume of a given mass of a gas is straightforwardly corresponding to its total temperature. 

V∝T

OR

V1/T1 = V2/T2 = constant


Combined Gas Law

This Law expresses that at a consistent volume, the pressure of a given mass of a gas is legitimately corresponding to its supreme temperature. This Law is a mix of Boyle's Law and Charles' Law. 

(P1V1)/T1 = (P2V2)/T2


Gay Lussac's Law

When volume is constant,

PαT

P1T1 = P2T2 = constant 


Avogadro's Law

Tests of various gases that contain a similar number of particles (any multifaceted nature, size, shape) involve a similar volume at a similar temperature and weight.

Vαn

V = kn

V/n = k = constant


Ideal Gas Equation

Ideal Gas Equation satisfies all three laws, i.e. Boyle's, Charles', and Avogadro's Law.

PV=nRT

PV=w/mRT

 OR 

P=d/mRT

 OR 

Pm=dRT 


Dalton's Law of Partial Pressures

The total pressure of a mixture of non-receptive gases at a consistent temperature and pressure is equivalent to the entirety of the individual halfway pressure of the gases.

Total Pressure = p1+p2+p3+...

p1=(n1RT)/v

p2=(n2RT)/v

p3=(n3RT)/v


Amagat's Law of Partial Volume

V = v1+v2+v3+....


Average Molecular Mass of Gaseous Mixture

The average molecular mass of the gaseous mixture is the ratio of the total mass of the mixture to the total number of moles in the mixture.

(n1M1+n2M2+n3M3+..) / (n1+n2+n3+..)


Graham's Law of Diffusion and Effusion

Diffusion is the ability of gas to spread and possess the entire accessible volume regardless of different gases present in the holder. Effusion is the process by which a gas escapes from one office of a vessel through a little opening or a hole. The rate of diffusion is measured by the ratio of the volume diffused to the time taken. Here, r is proportional to the inverse of the square root of density.

R = V /T.

r1/r2 = P2/P1 = (M2/M1)½ 

FAQs on Important Gaseous State Formula For JEE

1. In Light of Gaseous State, Explain How Boyle's and Charles' Law Can Be Represented Graphically.

Boyle's Law can be represented graphically as a plot of P versus 1/V at a steady temperature for a fixed mass of gas would be a straight line going through the cause. A plot of P versus V at a steady temperature for a fixed mass of gas would be a rectangular hyperbola. A plot of P (or V ) versus PV at a consistent temperature for a fixed mass of a gas is a straight line corresponding to the PV axis. Charle's Law can be represented graphically as a plot of volume versus temperature. At  steady weight, it is a straight line cutting the temperature hub at - 273 C and the straight line intersects the Y-axis at the level of volume. For an unmistakable mass of the gas, a plot of V versus T (K) at consistent weight is a straight line going through the origin. 

2. Define the Kinetic Molecular Theory of Gas.

Gases are made of an enormous number of indistinguishable particles (iotas or atoms), which are little and completely hard circles. The genuine volume of the particles is irrelevant as a contrast with the space among them, and subsequently, they are considered as the point masses. Association between the particles is unimportant. Particles of a gas are consistently in steady and irregular movement, and the impact between them is versatile. The normal motor vitality of the particles of a gas is straightforwardly corresponding to the supreme temperature. The pressure of a gas is because of the crash between gas particles and dividers of the compartment.