Notes on Gaseous State | Grade 11 > Chemistry > Gaseous State | KULLABS.COM

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Gaseous state of matter:

The most mobile state of matter is the gaseous state having the following characteristics:

1. The gas molecules neither have fixed shape nor have fixed volume both being determined by the volume of the container.

2.The gas is highly compressible in nature because of large intermolecular space between them.

3.They have less density and less viscosity.

4.The volume of the gas greatly changes with the change in temperature and volume.

5 The gas molecules have the very less intermolecular force of attraction because of large intermolecular space.

Some measurable properties of gas:

i. Pressure

- It is defined as the force applied per unit area of the gas.

- It is denoted by 'P'

- It is measured by barometer for atmospheric pressure and manometer for the closed vessel.

- It is expressed in Pa (Nm-2) or bar or atm or torr or mmHg

- 1 Pa = 1 bar = 1 atm = 1 torr = 760 mmHg

ii. Volume

- It is defined as the space occupied by the gas molecules.

- It is denoted by V.

- It is measured by measuring the volume of the container.

- It is expressed in ml or cc or cm3 or dm3 or l or m3

- 1 ml = 1cc

1000 ml/1000 cc = 1liter l

iii. Temperature

- It is defined as the average kinetic energy of the gas.

- It is denoted by T.

- It is measured by using a thermometer.

- It is expressed in terms of 0C or K.

- t0C = (t + 273) K

- The difference in temperature is equal to both the units.

Example: t1 = 200C

t2 = 300C

Now,

t2 - t1 = 30 - 20 = 100C

t2 - t1 = (30 + 273) - (20 + 273) = 303 - 293 = 10 K

iv.Mass (Amount)

- It is defined as the number of moles present in the gas.

- It is denoted by 'n'.

Number of mole (n) =$$\frac{Given\: Mass}{Molecular \:mass}$$

Example:Calculate the number of mole present in 8gm of oxygen gas.

Number of mole (n) = $$\frac{Given\: Mass}{Molecular \:mass}$$

Here, Given mass = 8 gm

Molecular mass = 32gm

Then,

n = $$\frac{8}{32}$$ = 0.25

Example:Calculate the number of moles present in 3.6 gram of sulphur dioxide gas.

Mass of SO2 gas (m) = 3.6 gm.

Molar mass = 64

Number of mole (n)== $$\frac{Given\: Mass}{Molecular \:mass}$$

= $$\frac{3.6}{64}$$

= 0.056

Example:Calculate the mass of hydrogen gas present in 3.2 moles?

Number of mole (n) = 3.2

Molar mass of hydrogen = 2

Mass of hydrogen = ?

Now ,

Number of mole (n) = $$\frac{Given\: Mass}{Molecular \:mass}$$

or, 3.2 = $$\frac{Given\: Mass}2$$

or, Given mass=3.2 x 2

∴ Mass of hydrogen = 6.4 gm.

Gas laws: The laws which describe the macroscopic behavior of gram molecule are called the gas law. Boyle's law, Charles law, Avogadro's law, Dalton's law of partial pressure and Graham's law of diffusion are the important laws which describe the gaseous behavior in terms of pressure, temperature and volume of gas.

Let 'P and 'V' are the pressure and volume of the gas at constant temperature 'T', then from Boyle's law, V ∝ $$\frac{1}{P}$$ (at constant temperature)

V = k $$\frac{1}{P}$$, where 'k' is the proportionality constant

or, PV = k ------------------- (i)

From equation (i), Boyle's law can also be stated as the product of pressure and volume of gas is constant at constant temperature.

Let V1 be the volume of the gas at pressure P1 and V2 be the volume of the gas at pressure P2. Then, from Boyle's law, we can write

P1V1 = P2V2

or, = $$\frac{P_1}{P_2}$$ = $$\frac {V_2}{V_1}$$

Graphical verification of Boyle's law

1. Graph of V vs. P: As we know from Boyle's law, PV = constant, so a plot of P vs. V gives a hyperbolic cure as shown in the figure.

[Graph of P vs. V]

1. Graph of V vs $$\frac{1}{P}$$:

PV = constant

V = $$\frac{1}{P}$$,. constant

which is in the form y = mx

So, a plot of V vs $$\frac{1}{P}$$,gives a straight line passing through the origin as shown in the figure.

1. Graph of PV vs P: The plot of PV vs P (or V) gives a straight line passing parallel to the pressure axis.

Relation between pressure and density of the gas

We know that,

Density (p) = $$\frac {Mass}{Volume}$$

From Boyle's law,

PV = k

or, P. $$\frac{Mass}{Density}$$ = k

or, $$\frac{P}{d}$$ =$$\frac{k}{M}$$

For a particlar gas, the mass of the gas remains constant. There the value of$$\frac{k}{M}$$also gives a constant value. Then,$$\frac{p}{d}$$= constant

or, p = constant x d

or, p ∝ d

For particular gas, pressure is directly proportional to the density of gas. So, a plot of p vs d will give a straight line passing through the origin.

•   It is measured by barometer for atmospheric pressure and manometer for the closed vessel.
• P and 'V' are the pressure and volume of the gas at constant temperature 'T', then from Boyle's law, V ∝ $$\frac{1}{P}$$ (at constant temperature)
• pressure is directly proportional to the density of gas. So, a plot of p vs d will give a straight line passing through the origin.
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