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1, What is meant by Diffusion?** ** Diffusion is a process of gradual mixing of two substances kept in contact.

2. Define the Gas Laws?** ** The behavior of a gas under known conditions of pressure, volume and temperature is described by the laws known as Gas Laws.

3. What are the Standard Variables for Gas Laws?** **The physical behavior of gases can be described by the three standard variables:

( I ) volume ( v ) (ii ) Pressure ( p)

(iii) temperature ( T )

4. Define atmospheric pressure?** **Our planet is surrounded by a thick blanket of air known as atmosphere. The Pressure exerted by air ( present in the atmosphere) on the surface of the earth is called atmospheric pressure.

5. Define Boyle’s law?

The volume of a given mass of a dry gas is inversely proportional to its pressure it constant temperature.

6, Define Boyle’s law equation?** **Boyle’s law may also be stated as the product of the volume and pressure of a given mass of a dry gas at a constant temperature is constant.

7, Define Charles’s Law?** ** Pressure remaining constant, the volume of a given mass of a dry gas increases or decreases by ^{1}/ _{273} of its volume at 0^{o} C for each 1^{o}c increase or decrease in temperature respectively.

8. Graphical representation of Charles’s Laws?

The relationship between the volume and the temperature of a gas can be plotted on a graph; as shown in Fig. A straight line is obtained.

The general term isobar which meant at constant pressure, is assigned at these plots.

9, What are the significance of Charles’s law?** ** Since the volume of a given mass of a gas is directly decreases with increase in temperature. This is the reason why hot air is filled in balloons used for meteorological purposes.

10, Explain the Mathematical expression of Boyle’s Law?** ** Suppose a gas occupies a volume V_{1}

When its pressure is P_{1} then

or P_{1} V_{1} = K = constant

If V_{2} is the volume occupied when the pressure is P_{2} at the same temperature, then

∴ P_{1} V_{1} = P_{2} V_{2} = K, at constant temperature.

11, Explain the Graphical verification of Boyle’s law?** ** The law can be verified by plotting a graph.

( i ) V vs ^{1}/_{p}

(ii ) V vs P

( iii) PV vs P**( i ) V vs ^{1}/_{ P}:** A straight line passing through the origin is obtained.

(ii) V vs P : < hyperbolic curve in the first quadrant is obtained;

Not: The term isotherm ( meaning at constant temperature ) is used to describe these plots.

( iii ) PV vs P : a straight line is obtained Parallel to the pressure axis.

12, Solved examples: 1

( 1 ) a gas occupies 800 cm3 under 760 mm Hg pressure. Find under what pressure the gas will occupy 380 cm3, the temperature remaining constant.

** **( P_{1} = 760 mm Hg; V_{1} = 800 cm^{3};

P_{2} = ? mm Hg; V_{2} = 380 cm^{3})

By Boyle’s law P_{1}V_{1} = P_{2}V_{2}

Substituting the values,

760 x 800 = P_{2} x 380

The required pressure is 160.0 cm Hg.

13, Example 2.

A gas occupies 600 cm3 under a pressure of 700 mm Hg. Find under what pressure the volume of the gas will be reduced by 20 percent of its original volume, the temperature remaining constant throughout?

**Ans:**

Therefore the new volume of the gas = 600 – 120 = 480 cm^{3}

(P_{1} = 700 Hg; V_{1} = 600 cm^{3},

P_{2} = ? mm Hg; V_{2} = 480 cm^{3})

By Boyle’s law, P_{1} v_{1} = P_{2} V_{2}

Substituting the values,

700 x 600 = P_{2} x 480

The required Pressure is 875 mm Hg.

14, Example 3:

Two cylinders, both containing carbon dioxide, are connected together by a tube fitted with a tap. The capacity of one cylinder is 4 dm^{3} and that of the other 1 dm^{3}; the Pressure in the first cylinder is 560 mm Hg and that in the second 1000 mm Hg . what will be the final pressure in both cylinders on opening the tap, if the temperature remain it constant ?

** ** Total Volume of carbon dioxide ( after opening the tap ) = 4 + 1 = 5 dm^{3}

For the first cylinder

P_{1}V1 = P_{2} V_{2}

560 x 4 = P2 x 5

For the Second cylinder

P_{1}V_{1} = P_{2} V_{2}

1000 x 1 = P_{2} x 5

Ans: final pressure on opening tap

= 448 + 200 = 648 mm Hg.

15, Example 4:

The volume of a given mass of a gas with some Pieces of marble in a container at 760 mm of Pressure is 100ml. If the Pressure is changed to 1000 mm Hg, the new volume is 80 ml. Find out the volume – occupied by the marble Pieces, if the temperature remains constant?

** ** Let the volume occupied by the marble Pieces = V ml.

At 760 mm of Hg, the volume occupied the gas = ( 100 –V ) ml.

At 100 mm of Hg, the volume occupied by the gas = ( 80 – V ) ml

By Boyle’s law, P_{1} V_{1} = P_{2} V_{2}

Therefore 760 x ( 100 – v ) = 1000 x ( 80 – v )

or 24 V = 400 ml

or V = 16.6 KL.

The required Volume = 16. 6 ml.

16, Mathematical expression of Charles’s Law.

Let V_{o} be the volume of a fixed mass of gas at 0^{o}C, and let V be its volume at temperature t^{o}C at same pressure. The according to Charles’s law.

17, Charles’s law may be restated as :

The volume of a given a mass of a day gas is directly proportional to its absolute ( Kelvin ) temperature of the pressure is kept constant.

Suppose, a gas occupies V_{1} cm^{3} at T_{1} temperature and V_{2} cm^{3} T_{2} temperature respectively. Then, by Charles’s law.

This is called Charles’s law equation.

According to this equation

(i ) If the temperature is doubled, the volume would be doubled.

(ii) If it is reduced to one – half, the volume would also be reduced to one half.

18, Solved examples:

Example 5

120 cm3 of a gas is taken at 27.3 k.

The temperature is then raised to 0^{o}C. What is the new volume of the gas? The pressure is kept constant throughout.

Ans: V_{1} = 120 cm^{3}; T_{1} = 27.3k; V_{2} = ? cm^{3}

T_{2} = 273 K, Since 0^{o}C = 273 K

By Charles’s law

Ans: The gas would occupy a volume of 1200 cm^{3} at 0^{o}C.

19, Example 5

At what temperature will 500 cm3 of a gas measured occupy half its volume ? The pressure is kept constant.

Ans: Let the required temperature be t^{o}C

V_{1} = 5000 cm^{3}; T_{1} ( 273 + 20 ) k = 293 k

V_{2} = 250 cm^{3}; T_{2} ( 273 + t ) K

By Charles’s law

or 500 ( 273 + t ) + 250 x 293

or 2 (273 + t ) = 293

or 2 (273 + t ) = ^{293}/_{2} = 146.5

or t = 146.5 – 273 = -126.5^{o}C

Ans: The gas would occupy half it s volume at -126.5^{o}C.

20, Example . 7

The volume of a given mass of a gas at 15^{o}C is 100 cm^{3}. To What temperature should it be heated under the Same pressure. So that it could occupy a volume of 125 cm^{3}?

Ans: Let the required temperature be t^{0}C

V = 100 cm^{3}; T = ( 273 + 15 ) k = 288 K

V^{1} = 125 cm^{3}; T^{1}= ( 273 + t ) k

By Charles’s law,

The Kelvin temperature can be converted to Celsius temperature by subtracting 273 ie.

360 k = ( 360 – 273 )^{o}C = 87^{o}C.

Ans: The gas would occupy 125 cm^{3}. At 87^{o}C

21. Example 8

At what centigrade temperature will the volume of a gas at 0^{o}C triple itself if pressure remains constant?

Let the volume at 0^{o}C = VmL

V_{1} = VmL; V_{2} = 3V mL

T_{1} = 0^{o} or 73 k; T_{2} = ?

By Charles’s law,

22. Example 9

87 cm3 of moist nitrogen is measured at 9o c and 659 mm. Hg. Pressure. Find the volume of dry nitrogen at NTP. The Vapour Pressure of water at 9^{o}C is 9 mm Hg.

**Ans:** Pressure due to dry nitrogen alone

= 659 – 9 = 650 mm Hg

P_{1} = 650 mm; V_{1} = 87 cm^{3}; T_{1} = ( 273 + 9 ) K

P_{2} = 760 mm ; V_{2} = 7 cm^{3}; T_{2} = 273 k

By the gas equation,

The volume of dry nitrogen at N.T. P would be 72.03 cm^{3}.

23 .Example 10,

The given mass of a gas occupies 572 cm3 at 13^{o}C and 725 mm Hg Pressure. What will be its volume at 24^{o}C and 792 mm Hg Pressure ?

P_{1} = 725 mm Hg; V_{1} = 572 cm;

T_{1} = ( 73 + 13 ) K = 286 K

P_{2} = 792 mm Hg; V_{2} ? cm^{3};

T_{2} = ( 273 + 24 ) k = 297 k

Applying the gas equation;

The required volume of gas is 543. 75 cm^{3}.

24, One litre of a gas at 10^{o} C is heated until both its volume and pressure are tripled.

Find the new temperature

Initial condition Final condition

P_{1} = P P_{2} = 3P

V_{1} = 1 L V_{2} = 3L

T_{1} = ( 273 + 10 )K = 283 K; T2 = ?

From the gas equation,

Ans: The new temperature is 2274^{o}C.

25, Example 12

A gas is enclosed in a cylinder under S.T. P conditions. At what temperature does the volume of the enclosed gas become 1/6^{th} of its initial volume ( pressure remaining Constant )

Ans: Let the Initial volume be V_{1} , final volume = ^{V1}/_{6}

Initial temperature T_{1} = 273 K final temperature = t^{2} = ?

By Charles’s law,

Ans: The temperature, at which the volume of the enclosed gas is 1⁄6^{th} of its initial volume is - 27.5^{o}C.

26, Example 13. The pressure of a gas at S.T.P is doubled and the temperature is raised to 546 K. What is the final volume of the gas ?

P_{1} = 760 mm Hg P_{2} = 2 x 760 mm Hg

T_{1} = 273 K T_{2} = 546 K

V_{1} =V_{1} V_{2} = ?

Ans: The volume of the gas remains the same.

27, Example – 14

Oxygen gas is enclosed in 1 dm^{2} flask at 25^{o}C. calculate the pressure exerted by gas. If the molecular mass ( molar mass ) of any gas occupies 22.4 litres at S.T.P.** ** Oxygen (O_{2} ) is a diatomic gas. It molar mass is 16 x 2 = 32 g

So, 32 g of oxygen occupies 22.4 dm^{3} at S.T.P 16 g of oxygen will occupy ^{22.4} /_{23} x 16 = 11. 2 dm^{3}

Thus

P_{1} = 1atm P_{2} = 2 x 760 mm Hg

V_{1} =11. 2 dm^{3} V_{2} = 1 dm^{3}

T_{1} = 273 k T_{2} = 273 + 25 = 298

By gas equation.

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Std 9

ICSE/ISC