WBBSE Solutions For Class 10 Physical Science And Environment Chapter 2 Behaviour Of Gases

WBBSE Class 10 Physical Science Question Answer In English

Chapter 2 Behaviour Of Gases Very Short Answer Type Questions

Question 1. Write the ideal gas equation for one gram-mole gas.
Answer: PV = RT.

Question 2. What is the value of absolute temperature in Fahrenheit scale?
Answer: (-) 459.4° F

Question 3. What is the SI unit of pressure?
Answer: The SI unit of pressure is Pascal.

Question 4. How is an atmosphere of gas determined?
Answer: the atmosphere of gas pressure is determined by Barometer invented by Torricelli.

Question 5. Give a mathematical expression of Boyle’s law.
Answer: V x P = Constant.

Question 6. Give a mathematical expression of Charles’s law.
Answer: \(\frac{\mathrm{V}}{\mathrm{T}}=\text { Constant. }\)

Question 7. What is meant by ideal gases?
Answer: Gases which obey Boyle’s law and Charle’s law at all pressure and temperature.

Read And Learn More: WBBSE Solutions For Class 10 Physical Science And Environment

Question 8. What is the effect of the withdrawal of heat from a gas at constant volume?
Answer: The temperature of the gas falls.

Question 9. What is Torr?
Answer: Torr is the unit of pressure used at high vacuum. 1 Torr 1 mm of Hg.

Question 10. Give a main property a gas.
Answer: Diffusion is a main property of a gas.

WB Class 10 Physical Science Question Answer

Question 11. What is Loschmidt’s number?
Answer: The number of molecules present in 1 ml. of gas or vapour at STP is known as the Loschmidt number.

Question 12. Is Charle’s law applicable to liquids?
Answer: Charle’s law is not applicable to liquids.

Question 13. What will be the volume of an ideal gas at absolute gas?
Answer: The volume, will be zero.

Question 14. What is the cause of the temperature of the gas?
Answer: Kinetic energy is possessed by gaseous molecules.

Question 15. Why are gaseous molecules always in incessant motion?
Answer: Because gaseous molecules have no mutual force of attraction.

Question 16. How does the volume of a gas change for each degree Celsius rise of temperature at constant pressure?
Answer: \(\frac{1}{273}\) of the volume of the gas at 0°C increases.

Question 17. What will happen when a gas-filled balloon is heated?
Answer: The balloon will burst as the supplied heat increases the volume of the gas inside the balloon.

Question 18. What are the variable quantities in Charle’s law?
Answer: In Charle’s law, the variable quantities are temperature and volume.

Question 19. What are the variable quantities in Boyle’s law?
Answer: In Boyle’s law, the variable quantities are pressure and volume.

14. What are the constant quantities in Charle’s law?
Answer: In Charle’s law, the constant quantities are the mass of the gas and pressure.

Physics Class 10 WBBSE Chapter 2 Behaviour Of Gases Fill In The Blanks

Question 1. The volume of a gas is zero at ________ temperature.
Answer: Absolute

Question 2. According to the kinetic theory of gases, the kinetic energy of gases is directly proportional to its ________.
Answer: absolute on Kelvin temperature.

Question 3. The law which describes the relationship between the volume and temperature of a gas at constant pressure is called _______.
Answer: Charle’s law.

Question 4. The scale of temperature with – 273°C as zero is called _________.
Answer: Absolute or Kelvin temperature scale.

WB Class 10 Physical Science Question Answer

Question 5. According to the kinetic theory of gases, a gas consists of discrete particles called _________.
Answer: molecules.

Question 6. Real gases do not behave like ideal gases because they have intermolecular _______ and the molecules do not have negligible volume.
Answer: attraction.

Question 7. Absolute scale reading = Celsius scale reading + _________.
Answer: 273.

Question 8. 330K ________ °C.
Answer: 57.

Question 9. The pressure of a gas is determined by _________.
Answer: Manometer.

Question 10. The value of normal pressure in the SI system is ________ pascal.
Answer: 1.013 × 10⁵.

Question 11. The boiling point of water is ________ Kelvin.
Answer: 273.

Question 12. At ________ °C, the gas molecules become motionless.
Answer: (-) 273.

Question 13. At constant temperature, the volume of a given mass of gas varies ________ as its pressure.
Answer: Inversely.

Question 14. For a given mass of a gas or vapour at a constant temperature, pressure ________ = constant.
Answer: Volume.

Question 15. The law of pressure is known as _________.
Answer: Regnault’s law.

Question 16. Any real gas becomes _______ before it reaches absolute zero.
Answer: Liquefied.

Question 17. P₁ V₁ = P₂ V₂ is related to ________.
Answer: Boyle’s law.

Question 18. The relation between pressure, temperature and density of a gas is _________.
Answer: \(\frac{\mathrm{P}}{\mathrm{DT}}=\text { constant }\)

Question 19. The boiling point of water on the Kelvin scale is _________.
Answer: 373 K.

Class 10 Physical Science WBBSE

Question 20. A constant temperature for a given volume of gas the product of its pressure and its volume is ________.
Answer: Constant.

Question 21. The value of absolute zero is ________ on the Celsius scale.
Answer: -273°C.

Question 22. The equation of ideal gas for a gram-moles is _________.
Answer: PV = nRT

Question 23. The volume of a given mass of gas varies directly as ________ temperature at constant pressure.
Answer: absolute.

Question 24. Gases like H₂ , O₂ , N₂ , Cl₂, etc. Which abey Boyle’s law and Charle’s law at low pressure and _______ temperature are called real gases.
Answer: high.

Question 25. The constant quantities in Boyle’s law are _______ and temperature.
Answer: mass.

Question 26. Gaseous molecules have no mutual force of _________ So they are always in an incessant motion.
Answer: Attraction.

Question 27. If heat is supplied to a gas, its molecules move with more ________ So the temperature of the gas rises.
Answer: energy.

Question 28. The normal temperature is taken as 0°C or _______ K and the normal pressure is taken as 76cm mercury or 1 atmosphere.
Answer: 273.

Class 10 Physical Science WBBSE

Question 29. Attractive forces and size effects in a gas can be neglected at ________.
Answer: low pressure and high temperatures.

Question 30. P, V, and T stand for pressure, volume and temperature of a gas _______ will express Boyle’s law.
Answer: \(V \times \frac{1}{P} \text { when } T \text { is constant }\)

Question 31. The relation between Celsius and absolute scale is ________.
Answer: T = t + 273.

Question 32. Normal pressure is _________.
Answer: 1.013 106 dyne/cm².

Question 33. Gases often deviate from the ideal gas behaviour because their molecules __________.
Answer: have forces of attraction between them.

Question 34. In the ideal gas equation PV = nRT, the value of R depends upon _________.
Answer: unit of measurement.

Question 35. At constant mass and constant volume, the relation between pressure and temperature is _________.
Answer: Gay Lussac’s law.

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

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

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

• 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:

• 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:

1. Four assumptions of the kinetic theory of gases:
2. All gases are made of molecules moving randomly in all directions.
3. The size of a molecule is much smaller than the average separation between the molecules.
4. The molecules exert no force on each other or on the walls of the container except during collision.
5. 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:

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

WBBSE Class 10 Physical Science Question Answer In English

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:

 

 

 

 

 

 

 

Chapter 2 Behaviour Of Gases Mathematical Problem

Question 1. A gas is allowed to expand at a constant temperature from an initial volume of 200 ml to a final volume 200 ml. At the end of the expansion the pressure of the gas was found to be in I atmosphere, what was the initial pressure of the gas?
Answer:

V₁ = initial volume of the gas = 200ml

P₁ = initial pressure of the gas =? atm

V₂= final volume of the gas = 1200 ml

P₂  = final pressure of the gas = 1 atm.

From Boyle’s law,

P₁V₁ = P₂V₂

X x 200 = 1 x 1200

\(X=\frac{1200}{200}\)

= 6 atoms

The initial pressure of the gas is 6 atm.

Question 2. A certain volume of the gas was found to be at a pressure of 1000 mm of mercury. When the pressure was decreased by 500 mm the gas occupied a volume of 2000 Cm³?
Answer:

Calculate the initial volume occupied by the gas if the change was done at a constant temperature.

V₁ = Initial volume of the gas = ? Cm³

P₁ = Initial pressure of the gas = 100 mm.

V₂ = Final volume of the gas = 2000 Cm³

P₂ = Final pressure of the gas = 500 m

From Boyle’s law,

P₁V₁ = P₂V₂

1000 × 500 × 2000

\(x=\frac{500 \times 2000}{1000}\)

= 1000 Cm³

The initial volume occupied by the gas is 1000 Cm³.

Question 3. A vessel of volume 2000 cm3 contains 0.1 mol of oxygen and 0.2 mol of carbon dioxide. If the temperature of the mixture is 300 K. Find its pressure.
Answer:

We know, \(\mathrm{p}=\frac{\mathrm{nRT}}{\mathrm{V}}\)

The pressure due to oxygen is

\(p_1=\frac{(0.1 \mathrm{~mol})\left(8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)(300 \mathrm{~K})}{\left(2000 \times 10^{-6} \mathrm{~m}^{-3}\right)}\)

= 1.25 x 10⁵ pa

Similarly, the pressure due to carbon dioxide

P₂ = 2.50 x 10⁵ pa

The total pressure in the vessel

P = P₁ + P₂

= (1.25 + 2.50) x 10⁵ pa

=3.75 x 10⁵ pa