#### Topics

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 0o C for each 1oc 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 V1
When its pressure is P1  then or P1 V1 = K = constant
If V2 is the volume occupied when the pressure is P2 at the same temperature, then ∴   P1 V1 = P2 V2 = 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.

( P1 = 760 mm Hg; V1 = 800 cm3;
P2 = ? mm Hg; V2 = 380 cm3)
By Boyle’s law P1V1 = P2V2
Substituting the values,
760 x 800 = P2 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 cm3
(P1 = 700 Hg; V1 = 600 cm3,
P2 = ? mm Hg; V2 = 480 cm3)
By Boyle’s law, P1 v1 = P2 V2
Substituting the values,
700 x 600 = P2 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 dm3 and that of the other 1 dm3; 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 dm3
For the first cylinder
P1V1 = P2 V2
560 x 4 = P2 x 5 For the Second cylinder
P1V1 = P2 V2
1000 x 1 = P2 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, P1 V1 = P2 V2
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 Vo be the volume of a fixed mass of gas at 0oC, and let V be its volume at temperature toC 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 V1 cm3 at T1 temperature and V2 cm3 T2 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 0oC. What is the new volume of the gas? The pressure is kept constant throughout.

Ans: V1 = 120 cm3; T1 = 27.3k; V2 = ? cm3
T2 = 273 K, Since 0oC = 273 K
By Charles’s law Ans: The gas would occupy a volume of 1200 cm3 at 0oC.

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 toC
V1 = 5000 cm3; T1 ( 273 + 20 ) k = 293 k
V2 = 250 cm3; T2 ( 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 = -126.5oC
Ans: The gas would occupy half it s volume at -126.5oC.

20, Example . 7
The volume of a given mass of a gas at 15oC is 100 cm3. To What temperature should it be heated under the Same pressure. So that it could occupy a volume of 125 cm3?

Ans: Let the required temperature be t0C
V = 100 cm3;    T = ( 273 + 15 ) k = 288 K
V1 = 125 cm3;   T1= ( 273 + t ) k
By Charles’s law, The Kelvin temperature can be converted to Celsius temperature by subtracting 273 ie.
360 k = ( 360 – 273 )oC = 87oC.
Ans: The gas would occupy 125 cm3. At 87oC

21. Example 8
At what centigrade temperature will the volume of a gas at 0oC triple itself if pressure remains constant?
Let the volume at 0oC = VmL
V1 = VmL; V2 = 3V mL
T1 = 0o or 73 k; T2 = ?
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 9oC is 9 mm Hg.

Ans: Pressure due to dry nitrogen alone
= 659 – 9 = 650 mm Hg
P1 = 650 mm; V1 = 87 cm3; T1 = ( 273 + 9 ) K
P2 = 760 mm ; V2 = 7 cm3; T2 = 273 k
By the gas equation, The volume of dry nitrogen at N.T. P would be 72.03 cm3.

23 .Example 10,
The given mass of a gas occupies 572 cm3 at 13oC and 725 mm Hg Pressure. What will be its volume at 24oC and 792 mm Hg Pressure ?
P1 = 725 mm Hg; V1 = 572 cm;
T1 = ( 73 + 13 ) K = 286 K
P2 = 792 mm Hg; V2 ? cm3;
T2 = ( 273 + 24 ) k = 297 k
Applying the gas equation; The required volume of gas is 543. 75 cm3.

24, One litre of a gas at 10o C is heated until both its volume and pressure are tripled.
Find the new temperature
Initial condition                     Final condition
P1 = P                                     P2 = 3P
V1 = 1 L                                 V2 = 3L
T1 = ( 273 + 10 )K = 283       K; T2 = ?
From the gas equation, Ans: The new temperature is 2274oC.

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 1/6th of its initial volume ( pressure remaining Constant )

Ans: Let the Initial volume be V1 , final volume = V1/6
Initial temperature T1 = 273 K final temperature = t2 = ?
By Charles’s law, Ans: The temperature, at which the volume of the enclosed gas is 1⁄6th of its initial volume is - 27.5oC.

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 ?
P1 = 760 mm Hg                       P2 = 2 x 760 mm Hg
T1 = 273 K                               T2 = 546 K
V1 =V1 V2 = ? Ans: The volume of the gas remains the same.

27, Example – 14
Oxygen gas is enclosed in 1 dm2 flask at 25oC. calculate the pressure exerted by gas. If the molecular mass ( molar mass ) of any gas occupies 22.4 litres at S.T.P.
Oxygen (O2 ) is a diatomic gas. It molar mass is 16 x 2 = 32 g
So, 32 g of oxygen occupies 22.4 dm3 at S.T.P 16 g of oxygen will occupy 22.4 /23 x 16 = 11. 2 dm3
Thus

P1 = 1atm                                        P2 = 2 x 760 mm Hg
V1 =11. 2 dm3                                  V2 = 1 dm3
T1 = 273 k                                       T2 = 273 + 25 = 298
By gas equation. Paid Users Only!
Paid Users Only!
Paid Users Only!
Paid Users Only!
Paid Users Only!
Paid Users Only!
Paid Users Only!
Paid Users Only!
Paid Users Only!
Paid Users Only!
Paid Users Only!
Paid Users Only!
Paid Users Only!
Paid Users Only! 