WBBSE Class 10 Physical Science Chapter 2 Behaviour Of Gases Long Answer Questions

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour Of Gases Long Answer Type Questions

Question 1. Establish-relation between the pressures and density of a gas at a constant temperature.
Answer:

Relation between the pressures and density of a gas at a constant temperature

Let the volume of the mass m of gas at a constant temperature be V₁ when the pressure is P₁ and V₂ when the pressure is P₂.

From Boyle’s law, P₁ V₁ = P₂ V_{2 →}(1)

Let the densities of the gas at the pressure P₁ and P₂ be P₁ and P₂ respectively. Substituting in equation (1) we get,

\(\frac{P_1 m}{P_1}=\frac{P_2 m}{P_2}\)

\(\frac{P_1}{P_1}=\frac{P_2}{P_2}\) ⇒ Ρ α Ρ

The density of a gas at constant temperature is proportional to pressure.

Question 2. What are the characteristic properties of gases?
Answer:

Characteristic properties of gas:

1. Shape and Volume: Since particles of gases are not held in fixed positions and move freely, gases neither have definite shapes nor definite volumes.
2. Homo generous nature: All parts of a gas or a gaseous mixture have similar composition throughout.
3. Density: Due to the large separation of molecules, gases have large volumes and thus low density.
4. Compressibility: On increasing pressure gases can be readily compressed due to the presence of large empty spaces.
5. Random motion: The molecules or atoms of a gas are in a state of continuous zig-zag motion in all directions.
6. Pressure: Due to their random motion, the molecules of the gas collide on the walls of the container and thus exert a certain force on the walls of the container. The force per unit area is called the gas pressure.
7. Diffessure: Gases mix (diffuse) with each other freely due to the free movement of their molecules.
8. Liquefaction: On cooling and applying. Pressure and gases can be liquefied. However, gas has to be cooled below a certain characteristic temperature called critical temperature before it can be liquefied by the application of pressure alone.

Question 3. What are the postulates of the kinetic theory of gases?
Answer:

Postulates of Kinetic theory of gases :

1. All gases consist of a very large number of tiny particles called molecules that are in constant rapid motion.
2. The gas molecules are perfectly round, very hard and separated by large distances. Their actual volume is thus, negligible as compared to the total volume of the gas.
3. The collisions between the gas molecules are perfectly elastic, i.e. there is no loss of energy during these collisions.
4. The distance between gas molecules being very large, there is no effective force of attraction or repulsion between them.
5. The average kinetic energy of the gas molecules is directly proportional to the absolute temperature of the gas.
6. The gas molecules collide with one another and with the walls of the container. The pressure exerted by a gas is due to the bombardment of its molecules on the walls of the vessel.
7. The gas molecules move freely in all directions. Their speed and direction change continuously due to collisions among them. As a result, their motion becomes zig-zag or random.

WB Class 10 Physical Science Question Answer

Question 4. A sample of helium has a volume of 520 cm3 at 373 k. Calculates the temperature at which the volume will become 260 Cm³. Assume that the pressure is constant.
Answer:

Given

A sample of helium has a volume of 520 cm3 at 373 k.

According to Charle’s law,

 

 

 

 

 

Question 5. Find the volume of N, at 27°C at a pressure of 760 torr, if it occupies 40 mL at STP.
Answer:

Given

at 27°C at a pressure of 760 torr, if it occupies 40 mL at STP.

According to Charle’s law,

 

 

 

 

 

Question 6. A sample of oxygen has a volume of 880 mL and a pressure of 740 torr. What additional pressure is required to reduce the volume to 440mL?
Answer:

Given

A sample of oxygen has a volume of 880 mL and a pressure of 740 torr.

According to Boyle’s law,

 

 

 

 

 

Question 7. 7.0g of a gas at 300K and 1 atmospheric pressure occupies a volume of 4.1 litters. What is the molecular mass of the gas?
Answer:

Given

7.0g of a gas at 300K and 1 atmospheric pressure occupies a volume of 4.1 litters.

Applying the ideal gas equation,

 

 

 

 

 

 

Question 8. A steel tank contains carbon dioxide at 27°C and a pressure of 10.0 atm. Calculate the internal gas pressure when the tank and the gas are heated to 100°C.
Answer:

Given

A steel tank contains carbon dioxide at 27°C and a pressure of 10.0 atm.

According to Gay Lussac’s law,

 

 

 

 

 

Question 9. A bottle of Volume 1 litre contains gas at 10 atmospheric pressure. How many bottles each of a 1-litre capacity can be filled with the gas at 2 atmospheric pressure at constant temperature?
Answer:

Given

A bottle of Volume 1 litre contains gas at 10 atmospheric pressure

According to Boyle’s law

WB Class 10 Physical Science Question Answer

Question 10. The volume of a certain mass of gas at 400k and pressure 202600 pascal is 2 cubic metres. What would be the volume of the gas at 327°C and 152cm of mercury Pressure? [given 1.013 x 10⁵ Pascal 1 atmospheric pressure]
Answer:

Given

The volume of a certain mass of gas at 400k and pressure 202600 pascal is 2 cubic metres.

According to Charle’s law

Question 11. An iron cylinder contains helium at a pressure of 250 K pa at 300k. The cylinder can withstand a pressure of 1×10⁶ pa. The room in which the cylinder is placed catches fire. Predict whether the cylinder will blow up before it melts or not. (M.P. of the cylinder 1800K)
Answer:

Given

An iron cylinder contains helium at a pressure of 250 K pa at 300k. The cylinder can withstand a pressure of 1×10⁶ pa. The room in which the cylinder is placed catches fire.

According to Gay-Lussac’s law.

 

 

 

 

 

Question 12. A certain quantity of gas occupies a volume of 1000 cm3 at 760 mm and 27°C. Find the volume of the gas if the pressure and temperature are 1520 mm and 327°C.
Answer:

Given

A certain quantity of gas occupies a volume of 1000 cm3 at 760 mm and 27°C.

By combining Boyle’s low and Charle’s low we get.

 

 

 

 

 

Question 13. A gas having a temperature 0°C is heated so that its pressure and volume are doubled. What will be the final temperature of the gas?
Answer:

Given

A gas having a temperature 0°C is heated so that its pressure and volume are doubled.

By combining Boyle’s low and Charle’s low we get

 

 

 

 

 

Question 14. 10g of Oxygen are introduced in a vessel of 5 lit capacity at 27°C calculate the pressure of the gas in the atmosphere in the container.
Answer:

Given

10g of Oxygen are introduced in a vessel of 5 lit capacity at 27°C

Applying the ideal gas equation,

 

 

 

 

 

Question 15. The volume of 1 gram mole of oxygen gas is 22400 ml at 760 mm. Calculate the temperature of the gas (Given R = 0.082L-atm mol^-1 K^-1)
Answer:

Given

The volume of 1 gram mole of oxygen gas is 22400 ml at 760 mm.

We know from the ideal gas equation for 1 mole of an ideal gas.

 

 

 

 

 

 

Question 16. Establish the equation:

\(V-t=V_0\left(1-\frac{t}{273}\right)\)
Answer:

Let, Vo denote the volume of a given mass of a gas at 0°C and Vt be the volume of the same mass of the gas at t°C, the pressure in both cases is the same.

From Charle’s law,

The increase in volume of the gas = \(\frac{\text { Volume at } 0^{\circ} \mathrm{C}}{273} \times \text { rise in temperature }\)

∴ \(V_t=V_0+\frac{V_0}{273} \times t\)

\(V_t=V_0\left(1-\frac{t}{273}\right)\)

If we consider a temperature lower than 0ºC, say tºC, then the volume of gas is \(v_{-1}=V_0\left(1+\frac{t}{273}\right)\)

WBBSE Class 10 Physical Science Solutions

Question 17. Establish: \(\frac{P}{T}=a\)constant for pressure law.
Answer:

If the pressure of a given mass of a gas at temperature t°c and the by P and P’ respectively, then volume remains constant.

We have Pressure law,

\(P=P o\left(1+\frac{t}{273}\right)=\frac{P o T}{273}\) \(p^{\prime}=P_0\left(1+\frac{t^{\prime}}{273}\right)=\frac{P_{o T}}{273}\)

Po = pressure of the gas.

TK = absolute temperature corresponding to the temperature t°c.

T1K= absolute temperature to the temperature t1 °C

\(\frac{P}{P^{\prime}}=\frac{T}{T^{\prime}} \)  ⇒ \(\frac{P}{T}=\frac{P^{\prime}}{T^{\prime}}\)

⇒ \(\frac{P}{T}\)= Constant when V = constant

∴ P α T

Question 18. Establish the relation PV KT for an ideal gas.
Answer:

Let, P = pressure of a gas

V = Volume of a gas

T= absolute temperature

From Boyle’s law,

\(V \propto \frac{1}{P}\) (T = Constant)

From Charle’s law,

V α T (P= constant)

When P and T both vary

\(V \propto \frac{T}{P}\)

⇒ \(V=K \frac{T}{P}\)

\(\frac{P V}{T}=K\)

PV = KT

Question 19. What is the general equation is obtained by combining Boyle’s law and Charle’s law. Define-universal gas constant. The general equation is obtained by combining Boyle’s law and Charles’s law.
Answer:

\(\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}\)

P₁ = Pressure of first gas

P₂ = pressure of second gas

V₁ = Volume of the first gas

V₂ = Volume of the second gas

T₁ = temperature of the first gas

T₂ = temperature of the second gas

Universal gas constant: One gram molecule of all gases occupy the same volume under identical conditions of temperature and pressure the volume of R is the same for all gases R is known as the universal gas constant.

WBBSE Class 10 Physical Science Solutions

Question 20. Establish a Relation between the pressure, temperature and density of a gas.
Answer:

Relation between the pressure, temperature and density of a gas

Let, P = Pressure of a gas

V = Volume of a gas

T = temperature of a gas

P= density of a gas

m = mass of a gas

M = Molecular weight of the gas

From the ideal gas equation,

\(\frac{P V}{T}=\frac{m}{M} R\) [R= universal gas constant]

\(\frac{\mathrm{PV}}{\mathrm{mT}}=\frac{\mathrm{R}}{\mathrm{M}}\) = Constant

But, \(\mathbf{P}=\frac{\mathbf{M}}{\mathbf{V}}\)

\(\frac{\mathbf{P}}{\mathrm{PT}}\)

 

Question 21. Why the absolute zero of temperature is called absolute?
Answer:

• We have seen that the concept of absolute zero of temperature is regardless of the kind, amount and initial pressure or volume of the gas. It is not possible to attain a temperature lower than this.
• But in other scales of temperature in the Celsius scale, the concept of 0°C depends on the behaviour of a particular material such as water. Temperature lower than 0°C are possible.
• So, the concept of absolute zero is more universal and more fundamental. Hence it is termed absolute.

Question 22. Explain the relation between the pressure-volume of a gas-When the temperature is kept constant.
Answer:

Relation between the pressure-volume of a gas-When the temperature is kept constant

• Pressure is increased.
• When the temperature of a fixed mass of a gas is kept constant and the pressure is increased systematically the volume correspondingly increases.
• Pressure is decreased.
• When the temperature of a fixed mass of a gas is kept constant and the pressure is decreased systematically the volume correspondingly increases.

Question 23. Explain the relation between temperature and volume of gas When the pressure is kept constant.
Answer:

Relation between temperature and volume of gas When the pressure is kept constant

• Temperature increased
• When the pressure of a fixed mass of gas is kept constant and the temperature of the gas is increased systematically the volume correspondingly increases.
• Temperature decreased:
• When the pressure of a fixed mass of gas is kept constant and the temperature of the gas is decreased slowly the volume correspondingly decreases.

Question 24. Write four assumptions of the kinetic theory of gases.
Answer:

Four assumptions of the kinetic theory of gases:

1. All gases are made of molecules moving randomly in all directions.
2. The size of a molecule is much smaller than the average separation between the molecules.
3. The molecules exert no force on each other or on the walls of the container except during collision.
4. The molecules obey Newton’s Laws of Motion.

Question 25. Establish: Vr ms = \(\sqrt{\frac{3 \mathrm{PV}}{\mathrm{M}}}\)
Answer:

The square root of mean square speed is called the root mean square speed or rms speed.

\(\mathrm{Vrms}=\sqrt{\Sigma v^2 / N}\)

V^{2 = (Vrms)}

We know that, P = \(\frac{1}{3} \mathrm{PV}^2 \mathrm{rms}\)

\(\text { Vrms }=\sqrt{\frac{3 p}{p}}=\sqrt{\frac{3 P V}{M}}\)

WBBSE Class 10 Physical Science Solutions

Question 26. Explain briefly the translational kinetic energy of a gas.
Answer:

The total translational kinetic energy of all the molecules of the gas is

\(K=\Sigma \frac{1}{2} m V^2\)

= \(\frac{1}{2} m \mathrm{~N} \frac{\Sigma \mathrm{V}^2}{\mathrm{~N}}\)

= \(\frac{1}{2} \mathrm{Mv}^2 \mathrm{rms}\)

The Average kinetic energy of a molecule

\(\frac{K}{N}=\frac{1}{2} \frac{M}{N} V^2 \mathrm{rms}\)

We know, \(\mathrm{pV}=\frac{1}{2} \mathrm{MV}^{-2}\) → (1)

From equation → (1)

\(\mathrm{pV}=\frac{2}{3} \frac{1}{2} \mathrm{Mv}^2 \mathrm{rms}\) \(p V=\frac{2}{3} K\) \(K=\frac{3}{2} p V\)

 

Question 27. State Avogadro’s law
Answer:

Avogadro’s law: At the same temperature and pressure, equal volumes of all gases contain equal numbers of molecules. This is known as Avogadro’s law.

Question 28. Establish Avogadro’s law.
Answer:

Avogadro’s law

Consider equal volumes of two gases kept at the same pressure and temperature.

m₁ = mass of a molecule of the first gas

m₂ = mass of a molecule of the second gas

N₁ = number of molecules of the first gas

N₂ = number of molecules of the second gas

P = Common pressure of the two gases

V = Common pressure of the two gases

∴ \(p V=\frac{1}{3} N_1 m_1 V_1^2\)

\(P V=\frac{1}{3} N_2 m_2 V_2\)

_{V1 V2} = rms speeds of the molecules of the first and second gas

\(N_1 m_1 V_1^2=N_2 m_2 V_2^2\) → (1)

As the temperature of the gases is the same, the average kinetic energy of the molecules is the same for the two gases.

\(\frac{1}{2} m_1 v_i==\frac{1}{2} m_2 v_2=\) → (2)

From, (1) and (2)

N₁ = N₂ Avogadro’s law

Question 29. Establish the relation: Dalton’s law of partial pressure: P=P, +P,+P, +……..
Answer:

In kinetic energy, the pressure exerted by a gas on the walls of a container is due to the collisions of the molecules of the walls.

The total force on the wall is the sum of the forces exerted by the individual molecules.

Suppose there are N₁ molecules of gas

1, N₂ molecules of gas 2 in the mixture.

The force on a wall of surface area A,

F= force by N₁ molecules of gas 1+ force by N₂ molecules of gas 2 + ….. = F+F+…

The pressure, \(P=\frac{F_1}{A}+\frac{F_2}{A}+\ldots \ldots\)

If the first gas alone is kept in the container, its N₁ molecules will exert a force F₁ on the wall.

If the pressure in this case P₁ , P₁ = \(\frac{F_1}{A}\)

Similar is the case for other gases Thus, P = P₁+P₂+P₃ +…

Question 30. Prive : \(\frac{r_1}{r_2}=\sqrt{\frac{p_2}{p_1}}\) for Graham’s law of diffusion.
Answer:

The rate of diffusion is proportional to the rms speed of the molecules of the gas.

If r₁ and r₂ be the rates of diffusion of the two gases,

\(\frac{r_1}{r_2}=\frac{V_1, r m s}{V_2, r m s}\) → (1)

\(\text { Vrms }=\sqrt{\frac{3 p}{P}}\)

If the pressure of the two gases is the same

\(\frac{\mathrm{V}_1, \mathrm{rms}}{\mathrm{V}_2 \mathrm{rms}}=\sqrt{\frac{\mathrm{p}_2}{\mathrm{p}_1}}\)

From equation → (1),

\(\frac{r_1}{r_2}=\sqrt{\frac{p_2}{p_1}}\) ..Graham’s law of diffusion

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Question 31. Prove Ideal gas equation ⇒ PV = nRT
Answer:

Consider a sample of an ideal gas at pressure P, volume V and temperature T.

Let, m = the mass of each molecule

V = rms speed of the molecule

Vtr = rms speed of the gas at the triple point 273. 16K

R = NaK = universal point

= universal gas point

R = 8.314 J K^-1 mol^-1

This is known as the equation of state of an ideal gas.

Question 32. Write the mathematical form of Max Well’s speed distribution law.
Answer:

The mathematical form of Maxwell’s speed distribution law :

\(d N=r \pi N\left(\frac{m}{2 \pi K T}\right)^{\frac{3}{2}} V^2 e^{-m V^2 / 2 K T}\)

 

Question 33. Write the mathematical expression of Vander Waal’s equation.
Answer:

The mathematical expression of the van der Waals equation.

\(\left(P+\frac{a}{V^2}\right)(V-b)=n R T\)

[a, b = constant

a = average force of attraction between the molecules

b = total volume of molecules]

1. A given sample of a substance has a number of parameters which can be physically measured. When these parameters are uniquely specified. So we say that the thermodynamic state of the system is specified.

We know. \(p V=\frac{1}{3} \mathrm{NmV}^2\) →(1)

\(\mathrm{T}=\left(\frac{273.16 \mathrm{k}}{\mathrm{V}^2 \mathrm{tr}}\right) \mathrm{V}^2\) \(V^2=\left(\frac{V_2 {tr}}{273.16 k}\right) T\)

Putting this Expression for V² in q^{n →}(1)

\(\mathrm{pV}=\mathrm{N}\left(\frac{1}{3} \frac{\mathrm{mV}^2 \mathrm{tr}}{273.16 \mathrm{k}}\right) \mathrm{T}\)

\(\frac{1}{2} \mathrm{mV}_{\mathrm{tr}}^2\) = average kinetic energy of a molecule at the triple point 273.16K.

As the average kinetic energy of a molecule is the same for all gases at a fixed temperature, mV2 tr is a universal constant. Accordingly. The quantity in the bracket in the equation.

2. above is also a universal constant writing this constant as a k equation.

3. becomes.

pV = NKT → (3)

The Universal constant K = Boltzmann Constant The value of K = 1.38 x 10^-23 JK^-1

If the gas contains n moles, the number of molecules is N = n^X_A

(NA 6.02 x 10²³mol^-1)

NA Avogadro’s Constant

Using eq (3) becomes p = x^N_A KT

PV = nRT → (4)

Question 34. Draw the graph Pressure (P) and volume (V) at a constant temperature.
Answer: