Frank Solutions for Chapter 2 Study of Gas Laws Class 9 Chemistry ICSE
1. Write the name and symbol of each state variable of a gas.
Answer
An ideal gas can be characterized by three state variables:
- Absolute pressure (P).
- Volume (V), and
- Absolute temperature (T).
2. State Boyle’s law.
Answer
Boyle’s law states that “At a constant temperature the volume of a fixed mass of gas is inversely proportional to its pressure.
V ∝ 1/P (At constant temperature)
3. Deduce a relationship between volume and pressure for a given mass of a gas at a constant temperature.
Answer
At constant temperature, volume of a given mass of a gas is inversely proportional to its pressure.
V ∝ 1/P
∴ V = K/P
Where, K is the constant of proportionality.
4. State Charles’ law.
Answer
Charles’ law states that “At constant pressure, the volume of a given mass of a dry gas is directly proportional to its absolute temperature (Kelvin).
V ∝ T (At constant pressure)
5. Write the value of Kelvin zero.
Answer
Kelvin zero is -273.15 ° C.
6. Write the mathematical statements of Boyles’ law and Charles; law. Use these statements to derive the combined gas equation.
Answer
According to Boyle’s law,
V α 1/P (At constant temperature)
Or PV = K …(1)
According to Charles’ law’
V ∝ T (At constant pressure)
Or V/T = K …(2)
On combining the above laws,
PV/T = K …(3)
Where, K is a constant.
Let us suppose that, for a given mass of a gas, the initial pressure, volume and temperature are P1, V1 and T1 which changes to P2, V2 and T2 respectively. Then
P1V1/T1 = K and P2V2/T2 = K
Therefore, P1V1/T1 = P2V2/T2 …(4)
Equation (4) is called combined gas law equation.
7. Define STP.
Answer
The standard temperature and pressure (STP by general convention are 0°C (273 K) and 1 atm (760 mm Hg)
8. (a) Write the value of standard temperature in (i) °C and (ii) K
(b) Write the value of standard pressure in (i) atm, (ii) mm Hg, (iii) cm Hg, (iv) torr
Answer
(a) The value of standard temperature is (i) 0°C and (ii) 273 K
(b) The value of standard pressure in (i) 1 atm, (ii) mm Hg, (iii) cm Hg, (iv) torr
9. A sample of gas occupies 45.6 cm3 at 240 mm Hg pressure and 25°C. What volume will it occupy at 720 mm Hg?
Answer
P1 = 240 mm Hg, P2 = 720 mm Hg
V1 = 45.6 cm3, V2 = ?
According to Boyle’s law, at constant temperature,
P1V1 = P2V2
⇒ 240 × 45.6 = 720 × V2
⇒ V2 = (240 × 45.6)/720
= 10944/720
= 15.2 cm3
10. A known amount of gas was enclosed in a vessel at a constant temperature and its volumes were measured by changing the pressure. In such an experiment, the following data were collected.
Answer
(i)
11. From the equation calculate the volume of ammonia gas formed when 6 litres of hydrogen react with excess nitrogen, all volumes being measured at STP.
Answer
Number of moles of ammonia gas 6/22.4 = 0.26 moles
3 moles of hydrogen form = 2 mole of ammonia
0.26 moles of hydrogen form = (2 × 0.26)/3 = 0.17 moles of ammonia
Volume of ammonia gas formed = 0.17 × 22.4 = 3.8 litres.
12. When stating the volume of a gas the pressure and temperature should also be given. Why ?
Answer
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.
13. Is it possible to change the temperature and pressure of a fixed mass of a gas without changing its volume ? Explain your answer.
Answer
Yes, it is possible to change the temperature and pressure of a fixed mass of a gas without changing its volume. For a given fixed mass of gas, number of moles of gas (i.e. n) remains constant. If the pressure and temperature of a given mass of a gas are changed simultaneously such that ratio of T and P remains constant, then volume will remain unchanged.
PV = nRT
⇒ V = nRT/P (where R is constant)
14. Explain briefly:
(a) The Kelvin scale has been adopted for chemical calculation.
(b) The absolute zero is a theoretical concept.
(c) When stating the volume of a gas, the pressure and temperature should also be given.
Answer
(a) Volume of a gas would be reduced to zero at 0 K (-273° C). 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.
15. 4 litres of CO2 is kept at 27°C. Calculate its volume if the volume is lowered to 150K at the same pressure.
Answer
Given:
V1 = 4 litres; V2 = ?
T1 = 27°C = 273 + 27 = 300K; T2 = 150K
According to Charles’ law, at constant pressure,
V1/T1 = V2/T2;
⇒ 4/300 = V2/150;
∴ V2 = (4 × 150)/300
= 2 litres
16. A gas cylinder containing cooking gas can withstand a pressure of 14.9 atmosphere. The pressure gauge of cylinder indicates 12 atmosphere at 27° C. Due to sudden fire in the building, its temperature starts rising. At what temperature will the cylinder explode?
Answer
Given;
P1 = 12 atm ; P2 = 14.9 atm
T1 = 27° C = 273 + 27 = 300°K; T2 =?
P1/T1 = P2/T2;
⇒ 12/300 = 14.9/T2 ;
⇒ T2 = (14.9 × 300)/12
= 372.5K
= 372.5 K
= 372.5 – 273
= 99.5° C
17. At 0°C and 760 mm Hg pressure, a gas occupies a volume of 100 cm3. The Kelvin temperature of the gas is increased by one fifth while the pressure is increased by one and a half times. Calculate the final volume of the gas.
Answer
Given;
P1 = 760 mm Hg; P2 = 3/2 × 760 = 1140 mm Hg
V1 = 100 cm3; V2 = ?
T1 = O°C = 273 K ; T2 = 6/5 × 273 = 327.6 K
Now,
P1V1/T1 = P2V2/T2;
⇒ (760 × 100)/273 = (1140 × V2)/327.6;
⇒ V2 = (760 × 100 × 327.6)/(273 × 1140)
= 80 litres
18. Which of the following statements are true and which are false?
(a) At constant pressure if a sample of a gas is heated from 0°C to 273°C, the volume will be double.
(b) At constant pressure if a sample of a gas is heated from 475°C to 1273°C, the volume will be double.
(c) Boyle studied the variation of the volume of a gas with pressure and found that as the pressure increases, so does the volume.
(d) A gas exerts pressure at the bottom but not on the walls of container.
(e) The P-V graph of a fixed mass of a gas at constant temperature is a straight line.
Answer
(a) True
(b) False
(c) False
(d) False
(e) False
19. How can you explain Boyle’s law and Charles’ law on the basis of the kinetic theory?
Answer
Boyle’s law-
From Kinetic theory of gases,
Charles’ Law
From gas law,
PV = NkT
At constant pressure and constant amount of gas, volume is directly proportional to the temperature which is the statement of Charles; law.
20. A sample of methane occupies 320 ml at 47°C. To what temperature (in degree celsius) will it have to be heated to occupy 450 ml at the same pressure?
Answer
Given;
V1 = 320 ml; V2 = 450 ml
T1 = 47°C = 47 + 273 = 320 K ; T2 = ?
V1/T1 = V2/T2;
⇒ 320/320 = 450/T2;
⇒ T2 = 450 K
= (450 – 273) °C
= 177°C
21. Give an experiment to show the pressure volume relationship of gas.
Answer
We trap a definite quantity of air in the closed vessel. At any point, the pressure on the air is equal to the atmosphere pressure plus the pressure due to 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 portion by which the volume decreases or increases is the same by which the pressure increases or decreases.
22. The molecular theory accounts for the pressure exerted by a gas in a closed vessel as a result of the gas molecules striking against the wall of the vessel. How will the pressure change if:
(a) The temperature is doubled keeping the volume constant,
(b) The volume is halved, keeping the temperature constant?
Answer
(a) Pressure will also be doubled.
(b) Pressure will be double.
23. Fill in the blanks:
(a) The melting point of the ice is ………..Kelvin.
(b) The temperature on the Kelvin scale at which molecular motion completely ceases is called………….
(c) The average kinetic energy of the molecule of a gas is proportional to the …………….
(d) If temperature is reduced to one half, ……….. would also reduced to one half.
Answer
(a) 273
(b) absolute zero
(c) absolute temperature
(d) the average kinetic energy
24. 2 litres of a gas is enclosed in a vessel when the pressure is 760 mm Hg. If the temperature remains constant, calculate the pressure when:
(a) the volume changes to 4 dm3,
(b) the volume is increased by one and half times.
Answer
(a) According to Boyles’ law,
P1 = 760 mm Hg; P2 = ?
V1 = 2 litres; V2 = 4 dm3 = 4 litres
⇒ P1V1 = P2V2
⇒ 760 × 2 = P2 × 4
⇒ P2 = (760 × 2)/4
= 380 mm Hg
(b) According to Boyles’ law,
P1 = 760 mm Hg; P2 = ?
V1 = 2 litre; V2 = 3/2 V1 = 1.5 × 2 = 3.0 L
Now,
P1V1 = P2V2
⇒ 760 × 2 = P2 × 3
⇒ P2 = (760 × 2)/3
= 506.66 mm Hg
25. Calculate the volume of dry air at STP that occupies 28 cm3 at 14°C and 750 mm Hg pressure when saturated with water vapour. The vapour pressure of water at 14°C is 12 mm Hg.
Answer
Given:
V1 = 28 cm3 ; P1 = 750 mm Hg
V2 = ?; P2 = 12 mm Hg
P1V1 = P2V2
⇒ 750 × 28 = 12 × V2
⇒ V2 = (750 × 28)/12
= 1750 cm3
26. A given amount of a gas X is confined in
a chamber at constant volume. When the chamber is immersed in a bath of melting
ice, the pressure of the gas is 100 cm Hg.
(a) What is the temperature, when the pressure
is 10 cm of Hg?
(b) What will be the pressure when the chamber
is brought to 100° C ?
Answer
(a)
Given:
P1
= 100 cm Hg ; T1 = 273 K
P2
= 10 cm Hg; T2 = ?
P1/T1
= P2/T2;
⇒ 100/273
= 10/T2;
(b)
Given:
P1
= 100 cm Hg ; T1 = 273 K
P2
= ? ; T2 = 100°C = 273 + 100
= 372 K
P1/T1
= P2/T2
⇒ 100/273
= P2/373;
P2
= (373 × 100)/273
P2 = 136.63 cm Hg
27. A fixed mass of a gas has a volume of 750 cm3 at -23°C and 800 mm pressure. Calculate the pressure for which the volume will be 720 cm3, the temperature being -3°C.
Answer
Given:
V1 = 750 cm3; T1 = - 23°C = 273 – 23 = 250 K; P1 = 800 mm Hg
V2 = 720 cm3; T2 = - 3°C = 273 – 3 = 270 K ; P2 = ?
(P1 × V1)/T1 = (P2 × V2)/T2;
⇒ (800 × 750)/250 = (P2 × 720)/270 ;
⇒ P2 = (270 × 800 × 750)/(250 × 720)
= 900 mm Hg
28. Why does the size of a weather balloon become larger and larger as it ascends into higher space?
Answer
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.
29. The pressure of a gas maintained in a steel tank is 20 atmospheres. If the pressure has to be increased by 20%, to what temperature must the tank be heated, if the initial temperature is 20°C ?
Answer
P1 = 20 atm ; T1 = 27°C = 273 + 27 = 300 K
P2 = 20 + 20 × 20/100 = 24 atm; T2 = ?
P1/T1 = P2/T2;
⇒ 20/300 = 24/T2;
⇒ T2 = (300 × 24)/20
= 360 K
= 360 – 273
= 87° C