States of Matter: Important Questions

Very Short Answer Questions [1 Mark Questions]

Ques. What are Van der waals forces?

Ans. Van der waals forces are weak forces of attraction found between the molecules of a substance. 

Ques. Define Boyle’s Law?

Ans. Boyle’s Law states that the pressure exerted by a gas is inversely proportional to the volume occupied by it, when temperature and no. of moles of the gas is constant. It can represented as;

P \(∝ \)1/V 

Ques. What is an Ideal Gas?

Ans. An ideal gas is the one which satisfies the ideal gas equation, that is, PV = nRT for any given value of temperature and pressure.

Ques. What is aqueous tension? 

Ans. The pressure exerted by the saturated water vapour on the walls of the container is called aqueous tension. 

Ques. Under what conditions do real gases show ideal behaviour? 

Ans. Under conditions of higher temperature and lower pressure, real gases tend to show ideal gas behaviour.

Ques. Why does the boiling point of a liquid rise when pressure is increased?

Ans. When the pressure exerted on a liquid is increased, more energy is required by the molecules to escape. Thus, the boiling point of the liquid increases.

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Short Answer Questions [2 Marks Questions]

Ques. The density of ice is lower than that of water. Give a reason. 

Ans. The structure of the ice is a cage-like 3D structure consisting of intermolecular hydrogen bonding. Upon melting, the hydrogen bonds break and the water molecules fill up the empty spaces. In the case of water, the molecules are more densely packed with each other than in ice. Hence, water has a higher density than ice.

Ques. What will be the molar volume of nitrogen and argon at 273.15 K and 1 atm?

Ans. 273.15 K is the standard value of temperature and 1 atm is considered as the standard pressure. At standard conditions of temperature and pressure (STP), the volume occupied by 1 mole of all the gases is 22.4 L. Hence, the molar volume of nitrogen and argon will be 22.4 L at 273.15 K and 1 atm.

Ques. What is the effect of hydrogen bonding on the viscosity of a liquid?

Ans. Hydrogen bonding increases the size of the molecules. With an increase in the size and mass of the molecule, the internal resistance of the molecule also increases and thus the viscosity of the liquid increases. 

Ques. Which are the two wrong assumptions in the kinetic theory of gases? 

Ans. The two wrong assumptions of kinetic theory of gases are:

1. There is no force of attraction between the molecules of the gas.
2. The volume of the molecules of the gas is negligibly small as compared to the total volume of the gas.

Ques. Derive the relationship between the density and molar mass of a gaseous substance.

Ans. According to the ideal gas equation,

PV = nRT

n/V = P/RT

Replacing n with m/M, we get

m/MV = P/RT (where, M = molar mass of the gas and m = mass of the gas)

Since density (d) = m/V

d/M = P/RT

M = dRT/P

Ques. Which factors affect the surface tension of a liquid?

Ans. The factors that affect the surface tension of a liquid are:

1. Nature of the liquid - The surface tension depends upon the intermolecular forces present between the molecules at the interface.
2. Temperature - As the temperature increases, surface tension decreases and becomes 0 at the critical temperature.

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Long Answer Questions [3 Marks Questions]

Ques. Give some important characteristics of gases.

Ans. The main characteristics of gases are:

1. They are highly compressible.
2. The density of gases is lower as compared to solids or liquids.
3. They do not possess fixed volume and shape.
4. They exert equal pressure in all directions.
5. They can expand to fill the container completely.
6. Two or more gases can mix easily and quickly with each other to form a homogeneous mixture.

Ques. Give reasons for the following:

• Liquid ammonia bottles are cooled before opening the seal.

• The tyre of an automobile is inflated at lesser pressure in summer than in winter.

Ans. a. The gas in the liquid ammonia bottle is kept under very high pressure. If the bottle is opened without cooling, the sudden decrease in the pressure will result in a large increase in the volume of the gas. Thus, the gas will come out of the bottle with great force. To avoid this, the bottle is cooled under running tap water in order to decrease its volume.

1. The pressure of the gas is directly proportional to the temperature. In summer, temperature is high and so the pressure will be high in the tyre. In such a high pressure, the tube may burst. Therefore, tires are inflated with lower pressure in summer than in winter.

Ques. What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm3 flask at 27°C? [NCERT]

Ans. p = n/V.RT = m/M.RT/V

PCH₄ = (3.2/16 mol) 0.0821 dm³atmK^-1mol^-1 300K / 9 dm³ = 0.55 atm

PCO₂ = (4.4/44 mol) 0.0821 dm³atmK^-1mol^-1 300K / 9 dm³ = 0.27 atm 

Ptotal = 0.55 + 0.27 = 0.82 atm

Ques. Calculate the total number of electrons present in 1.4g of dinitrogen gas. [NCERT]

Ans. Molecular mass of N₂ = 28 g

28 g of N₂ has no. of molecules = 6.022 10²³

1.4 g of N₂ has no. of molecules = 6.022 10²³ 1.4 g / 28 g = 3.011 10²² molecules

Atomic no. of nitrogen (N) = 7

1 molecule of N₂ has electrons = 7 2 = 14

3.011 10²² molecules of N₂ have electrons = 14 3.011 10²² = 4.215 10²³ electrons

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Very Long Answer Questions [5 Marks Questions]

Ques. At 25°C and 760 mmHg pressure a gas occupies 600 ml volume. What will be its pressure at a height where temperature is 10°C and volume of the gas is 640 ml? [NCERT]

Ans. P₁ = 760 mmHg 

V₁ = 600 ml

T₁ = 25 + 273 = 298 K

V₂ = 640 ml 

T₂ = 10 + 273 = 283 K

According to the combined gas law, 

P₁V₁/T₁ = P₂V₂/T₂ 

P₂ = P₁V₁T₂ / T₁V₂ 

P₂ = 760 mmHg 600 ml 283 K / 640 ml 298 K

P₂ = 676.6 mmHg 

Ques. One litre flask containing vapours of methyl alcohol (Mol. mass = 32) at a pressure of 1 atm and 25°C was evacuated till the final pressure was 10-3 mm. How many molecules of methyl alcohol were left in the flask?

Ans. P1 = 10^-3 mm

V₁ = 1000 cm³ 

T₁ = 298 K

Converting this volume to volume, at STP, where T₂ = 2/3 K and P₂ = 760 mm 

P₁V₁ / T₁ = P₂V₂ / T₂ 

10^-3 1000 / 298 = 760 V2 / 273

V2 = 1.205 10^-3 cm³ 

Now, 22400 cm³ at STP contains 6.02 10²³ molecules 

∴ 1.205 10^-3 cm³ at STP contains 

= 6.022 10²³ / 22400 1.205 10^-3 molecules

= 3.24 10¹⁶ molecules

Ques. Calculate the pressure exerted by 110 g of CO₂ in a vessel of 2L capacity at 37°C. Given that the van der waals constants are a = 3.59 L² atm mol-² and b = 0.0427 L mol^-1. Compare the value with the calculated value if the gas were considered ideal. 

Ans. According to Van der waals equation,

(P + an²/V²) (V - nb) = nRT

P = (nRT / V - nb) - (an²/V²)

Here n = 110/44 = 2.5 moles. Putting the given values, we get

P = [2.5 0.0821 310 / (2 - 2.5 0.0427)] - [3.59 2.5 / 2]

P = 33.61 atm - 5.61 atm = 28.0 atm

If the gas were considered an ideal gas, applying ideal gas equation

PV = nRT

Or P = nRT/V

∴ P = 2.5 0.0821 310 / 2 = 31.8 atm

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