FRANK Solutions for Class 9 Chemistry Chapter 2 - Study of Gas Laws

Gain conceptual knowledge with Frank Solutions for ICSE Class 9 Chemistry Chapter 2 Study of Gas Laws. Study Boyle’s Law and Charles’ Law by practising our textbook solutions. Also, go through questions and answers on standard temperature and pressure.

Practise problems on calculating the final volume of gas, the volume of dry air, and more. Study gas laws for your exam thoroughly with the ICSE Class 9 Chemistry Frank solutions and Selina solutions at TopperLearning. Also, browse through our learning materials such as sample paper solutions, video lessons, online practice tests and more to help you understand concepts better.

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Chapter 2 - Study of Gas Laws Exercise 21

Solution 4

Solution 5

Kelvin zero is -273.15^oC.

Solution 6

Solution 7

The standard temperature and pressure (STP) by general convention are 0^OC(273 K) and 1 atm(760 mm Hg).

Solution 8

(a) The value of standard temperature is (i) 0^oC and (ii) 273 K
(b)The value of standard pressure is (i) 1 atm, (ii) 760 mm of Hg, (iii)76 cm of Hg, (iv)760 torr

Solution 9

Solution 10

Solution 11

Solution 12

There is simultaneous effect of temperature and pressure changes on the volume of a given mass of a gas. So, when stating the volume of a gas, the pressure and temperature should also be given.

Solution 1

An ideal gas can be characterized by three state variables:
(i) Absolute pressure (P),
(ii) Volume (V), and
(iii) Absolute temperature (T).

Solution 2

Solution 3

Chapter 2 - Study of Gas Laws Exercise 22

Solution 19

Solution 27

Solution 20

Solution 28

As weather balloon go higher into the atmosphere, the air becomes less dense, so air pressure drops. Because of this, the air that is already inside the balloon expands to cope with the difference in pressure. The end result is that the balloon expands making it larger.

Solution 21

We trap a definite quantity of air in the closed vessel. At any point, the pressure on the air is equal to the atmospheric pressure plus the pressure due to the excess mercury column in the open end tube. By pouring mercury in the tube, we increase the pressure on the air and measure its volume under that pressure. We thus obtain a set of data for the volume of a fixed mass of air under different pressures.
For a given mass of air at constant temperature, the following observations are made-
(i) The volume of air decreases with increasing pressure and vice versa.
(ii) The proportion by which the volume decreases or increases is the same by which the pressure increases or decreases.

Solution 29

Solution 22

(a) Pressure will also be doubled.
(b) Pressure will be double.

Solution 13

Solution 23

(a) 273
(b) absolute zero
(c) absolute temperature
(d) the average kinetic energy

Solution 14

(a) Volume of a gas would be reduced to zero at 0 K (-273⁰C).All temperatures on the Kelvin scale are positive, so Kelvin scale has been adopted for chemical calculation.
(b) At absolute zero temperature, volume of a gas would be reduced to zero. Theoretically,this is the lowest temperature that can be reached. At this temperature all molecular motions cease. Thus, practically this temperature is impossible to attain because on cooling gases liquefy and Charles' law is no more applicable.
(c) According to combined gas law equation, there is simultaneous effect of temperature and pressure changes on the volume of a given mass of a gas. So, when stating the volume of a gas, the pressure and temperature should also be given.

Solution 15