# 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 Î± 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 P_{1}, V_{1} and T_{1} which changes to P_{2}, V_{2} and T_{2} respectively. Then

P_{1}V_{1}/T_{1} = K and P_{2}V_{2}/T_{2} = K

Therefore, P_{1}V_{1}/T_{1} = P_{2}V_{2}/T_{2} **…(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 cm ^{3} at 240 mm Hg pressure and 25°C. What volume will it occupy at 720 mm Hg? **

**Answer**

P_{1} = 240 mm Hg, P_{2} = 720 mm Hg

V_{1} = 45.6 cm^{3}, V_{2} = ?

According to Boyle’s law, at constant temperature,

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

⇒ 240 × 45.6 = 720 × V_{2 }

⇒ V_{2 }= (240 × 45.6)/720

= 10944/720

= 15.2 cm^{3}

**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. **

**Using the given data, make the plots of (i) V versus P (ii) V versus 1/P and (iii) PV versus P and explain that the gas obeys Boyle’s law.**

**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 CO _{2} is kept at 27°C. Calculate its volume if the volume is lowered to 150K at the same pressure. **

**Answer**

Given:

V_{1} = 4 litres; V_{2 }= ?

T_{1} = 27°C = 273 + 27 = 300K; T_{2} = 150K

According to Charles’ law, at constant pressure,

V_{1}/T_{1} = V_{2}/T_{2};

⇒ 4/300 = V_{2}/150;

∴ V_{2} = (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;

P_{1} = 12 atm ; P_{2} = 14.9 atm

T_{1} = 27° C = 273 + 27 = 300°K; T_{2} =?

P_{1}/T_{1} = P_{2}/T_{2};

⇒ 12/300 = 14.9/T_{2} ;

⇒ T_{2} = (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 cm ^{3}. 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;

P_{1} = 760 mm Hg; P_{2} = 3/2 × 760 = 1140 mm Hg

V_{1} = 100 cm^{3}; V_{2} = ?

T_{1} = O**°**C = 273 K ; T_{2} = 6/5** **× 273 = 327.6 K

Now,

P_{1}V_{1}/T_{1} = P_{2}V_{2}/T_{2};

⇒ (760 × 100)/273 = (1140 × V_{2})/327.6;

⇒ V_{2} = (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;

V_{1} = 320 ml; V_{2} = 450 ml

T_{1} = 47°C = 47 + 273 = 320 K ; T_{2} = ?

V_{1}/T_{1} = V_{2}/T_{2};

⇒ 320/320 = 450/T_{2};

⇒ T_{2} = 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 dm ^{3}, **

**(b) the volume is increased by one and half times. **

**Answer**

(a) According to Boyles’ law,

P_{1} = 760 mm Hg; P_{2} = ?

V_{1} = 2 litres; V_{2} = 4 dm^{3} = 4 litres

⇒ P_{1}V_{1} = P_{2}V_{2}

⇒ 760 × 2 = P_{2} × 4

⇒ P_{2} = (760 × 2)/4

= 380 mm Hg

(b) According to Boyles’ law,

P_{1} = 760 mm Hg; P_{2} = ?

V_{1} = 2 litre; V_{2} = 3/2 V_{1} = 1.5 × 2 = 3.0 L

Now,

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

⇒ 760 × 2 = P_{2} × 3

⇒ P_{2} = (760 × 2)/3

= 506.66 mm Hg

**25. Calculate the volume of dry air at STP that occupies 28 cm ^{3} 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:

V_{1 }= 28 cm^{3} ; P_{1} = 750 mm Hg

V_{2} = ?; P_{2} = 12 mm Hg

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

⇒ 750 × 28 = 12 × V_{2}

⇒ V_{2} = (750 × 28)/12

= 1750 cm^{3}

**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:

P_{1}
= 100 cm Hg ; T_{1} = 273 K

P_{2}
= 10 cm Hg; T_{2} = ?

P_{1}/T_{1}
= P_{2}/T_{2};

⇒ 100/273
= 10/T_{2};

(b)
Given:

P_{1}
= 100 cm Hg ; T_{1} = 273 K

P_{2}
= ? ; T_{2} = 100°C = 273 + 100
= 372 K

P_{1}/T_{1}
= P_{2}/T_{2}

⇒ 100/273
= P_{2}/373;

P_{2}
= (373 × 100)/273

P_{2} = 136.63 cm Hg

**27. A fixed mass of a gas has a volume of 750 cm ^{3} at -23°C and 800 mm pressure. Calculate the pressure for which the volume will be 720 cm^{3}, the temperature being -3°C. **

**Answer**

Given:

V_{1} = 750 cm^{3}; T_{1} = - 23°C = 273 – 23 = 250 K; P_{1} = 800 mm Hg

V_{2 }= 720 cm^{3}; T_{2} = - 3°C = 273 – 3 = 270 K ; P_{2} = ?

(P_{1} × V_{1})/T_{1} = (P_{2} × V_{2})/T_{2};

⇒ (800 × 750)/250 = (P_{2 }× 720)/270 ;

⇒ P_{2} = (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**

P_{1 }= 20 atm ; T_{1} = 27°C = 273 + 27 = 300 K

P_{2} = 20 + 20 × 20/100 = 24 atm; T_{2} = ?

P_{1}/T_{1} = P_{2}/T_{2};

⇒ 20/300 = 24/T_{2};

⇒ T_{2} = (300 × 24)/20

= 360 K

= 360 – 273

= 87° C