1st PUC Chemistry Question Bank Chapter 5 States of Matter

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Karnataka 1st PUC Chemistry Question Bank Chapter 5 States of Matter

Question 1.
Name the three physical states of matter.
Answer:
Solid, liquid and gas.

Question 2.
Give the factors which determine the physical state of matter.
Answer:

• Nature of intermolecular forces
• Molecular interactions
• Thermal energy

Intermolecular Forces

Question 1.
What are intermolecular forces?
Answer:
The forces of attraction and repulsion between interacting particles (atoms ions and molecules) is called intermolecular forces.

Question 2.
What are van der Waals forces?
Answer:
The weak attractive forces between the molecules are known as van der Waals forces.

Question 3.
What are London forces or dispersion forces?
Answer:
The force of attraction between the induced momentary dipoles is called London or dispersion forces.
Note:
1. London forces are always attractive and interaction energy is inversely proportional to the sixth power of the distance between two interacting particles (i.e., \(\frac{1}{r^{6}}\) where r is the distance between two particles).

2. These forces are important only at short distances and their magnitude depends on the polarisibility of the particle.

Question 4.
What are dipole – dipole forces? Explain with an example.
Answer:
The attractive forces between positive end (δ⁺) of the one molecule and negative end (δ^–) of the other molecule is called dipole – dipole attractive forces.
Example: Hydrogen chloride molecules have dipoles depicted below.

The polar HCl molecules interact with neighbouring HCl molecules. This interaction is stronger than London forces but is weaker than ion-ion interaction because only partial charges are involved. The attractive force decreases with the increase of distance between the dipoles.

Note:

1. Dipole – dipole interaction energy is directly proportional to distance between polar molecules.
2. Dipole – dipole interaction energy between stationary polar molecules is proportional to  \(\frac{1}{r^{3}}\)
3. Dipole – dipole interaction energy between rotating polar molecules is proportional to \(\frac{1}{r^{6}}\) where r is the distance between polar molecules.

Question 5.
Explain dipole – induced dipole attractive forces with an illustration.
Answer:
The attractive force between the permanent polar molecules and induced polar molecules is called Dipole – Induced dipole attractive force.

Polar molecules induce polarity on non polar molecules in the opposite direction by polarizing the electron cloud. The phenomenon can be represented as follows.
Polarization

The interaction energy in this case is also proportional to \(\frac{1}{r^{6}}\) where r is the distance between the dipole and induced dipole.

Larger the size of non-polar molecule more is its polarizability. That is why dipole – induced dipole attractive force is more between the molecules of water and Neon than that between water and helium molecules.

Question 6.
In which type of molecules dipole – induced dipole forces operate?
Answer:
Polar molecules having permanent dipole and molecules having no dipole.

Question 7.
On what factors induced dipole moment depends?
Answer:

• Dipole moment present in the permanent dipole.
• Polarisability of the electrically neutral molecule.

Thermal Energy

Question 1.
What is thermal energy? How is it related to the temperature?
Ans:
The energy of a body arising from motion of its atoms or molecules is called thermal energy. Thermal energy is directly proportional to temperature of the substance.

Question 2.
What is thermal motion?
Answer:
Thermal energy is the measure of average kinetic energy of the particles of the matter and is responsible for movement of particles is called thermal motion.

Question 3.
Arrange solids, liquids and gases in the decreasing order of their thermal forces.
Answer:
Gas > liquid > solid.

Question 4.
Mention the two factors which decide the physical state of a substance.
Answer:

1. Thermal energy
2. Intermolecular forces

Intermolecular Forces vs Thermal Interactions

1. Three states of matter are the result of balance between intermolecular forces and the thermal energy of the molecules.
2. Predominant of intermolecular forces increases in the order : Gas → Liquid → Solid. Predominance of thermal energy increases in the order: Solid → liquid → Gas.

Question 1.
Arrange solids, liquids and gases in the increasing order of their intermolecular forces.
Answer:
Gas < liquid < solid

The Gaseous State

Question 1.
Give the characteristics of gases.
Answer:

• Gases are highly compressible.
• Gases exert pressure equally in all directions.
• Gases have lower density than liquids and solids.
• Gases do not have definite shape and volume.
• Gases mix completely in all proportions without any mechanical aid.

The Gas Laws

Question 1.
State Boyle’s law. Give its mathematical expressions.
Answer:
This law states that “at constant temperature, the pressure of a fixed amount of gas is inversely proportional to its volume”.
Mathematically, P ∝ \(\frac { 1 }{ v }\) at constant T …… (1)
or pV ∝ 1 or pV = constant
For 2 gases P₁V₁ = constant
P₂V₂ = constant
[∴ P₁V₁ = P₂V₂]
P₁V₁ = constant and p₂V₂ = constant
∴ P₁V₁ = p₂V₂ or \(\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}=\frac{\mathrm{V}_{2}}{\mathrm{V}_{1}}\)

Question 2.
Explain Boyle’s law graphically.
Answer:
Variation of volume of a given mass of gas with pressure at constant temperature can be represented can be represented graphically as shown below.

The graphs are called isotherms. From Figure it is clear that when pressure is increased by n times, volume
decreases by n times. From Boyle’s law, p = k \(.\frac{1}{V}.\) This equation is in the form of y = mx. This indicates that the plots of p vs \(\frac{1}{V}\) at constant temperatures must be straight lines passing through the origin. The slopes of these straight lines give the values of k at different temperatures. Since pV = k, the graph of pV vs p at constant temperature must be a horizontal line parallel to the x-axis.

Question 3.
Derive the relation between density and pressure of a gas using Boyle’s law.
Answer:
Gases are highly compressible because of empty space present between the gas molecules (i.e., occupy smaller space when a given mass of a gas is compressed). This means that density of a gas becomes more at high pressure.
By definition,
Density = \(\frac{\text { Mass }}{\text { Volume }}\) , Therefore, d = \(\frac{m}{V}\) where ‘m’ is the mass and ‘ V’ is the volume.
∴ V = \(\frac{m}{d}\)
According to Boyle’s law, V = k \(\frac{1}{p}\)
Substituting the value of ‘V’ in the above equation, we get, \(\frac{m}{d}=\frac{k}{p}\)
d = \(\left(\frac{m}{k}\right) p\)
d = k’p where k’ = constant.
∴ d α p
This shows that at constant temperature the density of a fixed mass of a gas is directly proportional to the pressure.

Question 4.
State Charle’s law. Give is mathematical expression.
Answer:
It states that “pressure remaining constant, the volume of a fixed mass of a gas is directly proportional to its absolute temperature”.
V ∝ T ………. at constant pressure
V = k₂T
\(\frac{v}{T}\) = K₂
Under two different conditions,
\(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\)
V₁T₂ = V₂T₁
This relation is also called Charles law.

Question 5.
Define Kelvin temperature scale or absolute temperature scale.
Answer:
The temperature scale in which t °C on a new scale is given by t°C + 273.15 and 0°C by T₀ = 273.15 K is called Kelvin temperature scale or absolute temperature scale.
Note: Kelvin temperature scale is also called as thermodynamic scale of temperature and is used in all scientific works.

Question 6.
What is absolute zero temperature?
Answer:
The lowest hypothetical or imaginary temperature (-273.15 °C) at which volume of any gas becomes zero is called absolute zero temperature.

Question 7.
Explain Charle’s law graphically.
Answer:
The variation in volume of a given mass of gas with temperature at constant pressure can be represented graphically. The graph is called an isobar. Isobars for a given mass of a gas at different pressures are given in the graph. On extending these isobars to zero volume, all of them concur at – 273.15°C on the x-axis. This also suggests that the lowest possible temperature is – 273.15°C. Later thermodynamically also it was proved to be so and – 273.15°C is called absolute zero of temperature.

Question 8.
Show that V₁ = V₀\(\left(\frac{273.15+t}{273.15}\right)\)
Answer:
Let the volume of a given amount of gas be V₀ at 0°C. The temperature is increased by t°C. Then the volume of the gas also increases to V_t. Experimental studies on gases by Charle’s has shown that V₁ and V₀ are related to each other as

Question 9.
State and explain Gay-Lussac’s law.
Answer:
It states that, “the pressure of a given mass of a gas is directly proportional to the absolute temperature at constant volume.”
Thus p α T at constant volume for a given mass of gas
or p = kT
where k is constant.
Therefore, we can write, \(\frac{p_{1}}{T_{1}}=\frac{p_{2}}{T_{2}}\), where p₁ and p₂ are the pressures of the gas at temperatures T₁ and T₂, respectively.

Graphical representation of the variation of pressure of a given mass of a gas with temperature at constant volume is called an isochore. Isochores for a given mass of a gas at different constant pressure are given in the graph.

Question 10.
State Avogadro’s law.
Answer:
This law states that “equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules”. If a gas contains n₁, molecules in volume V and gas B contains n₂ molecules in volume V at the same temperature and pressure, then according to the law n₁ = n₂. Avogadro’s law can also be stated as “the volume of a gas is directly proportional to its number of moles as long as the temperature and pressure are kept constant”.
[V α v] Where ‘n’ is the number of moles.
V = kn …… (1)
where ‘k’ is a constant.

Question 11.
Derive the relation between density and molar mass of gas Using Avogadro’s law.
Answer:
If the number of moles of gas is changed from n₁ to n₂ at constant T and P, we can write Avogadro’s Law as;
\(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\)
Number of moles of a gas(n) can be calculated as follows:
n = \(\frac{\mathrm{m}}{\mathrm{M}}\) where m is mass of the gas under investigation and M = molar mass. M
Therefore equation (1) can be written as
V = k\(\frac{\mathrm{m}}{\mathrm{M}}\)
M = k\(\frac{\mathrm{m}}{\mathrm{V}}\)
M = k₄d
M ∝ d
Where d is the density of gas. Thus at constant temperature and volume, higher is the molecular mass more is the density.

Numerical Problems

Question 1.
240 cm³ of air at a pressure of 100 kPa in a bicycle pump is compressed to a volume of 150 cm³. What is the pressure of the compressed air in the pump?
Answer:
p₁ × V₁ = p₂ × V₂ (From Boyle’s law)
p₂ = \(\frac{p_{1} V_{1}}{V_{2}}=\frac{100 \times 240}{150}\) = 160Kpa

Question 2.
10 m³ of butane gas at 1.2 atm was required to be stored at 6 atm pressure. To what volume must the gas be compressed to give the required storage pressure at constant temperature?
Answer:
p₁ × V₁ = p₂ × V₂ (From Boyle’s law)
V₂ = \(\frac{p_{1} V_{1}}{p_{2}}=\frac{1 \cdot 2 \times 10}{6}\) = 2m³

Question 3.
The pressure exerted by a gas in a sealed container is 100 kPa at 17°C. It was found that the container might leak if the internal pressure exceeds 120 kPa. Assuming constant volume, at what temperature in °C will the container start to leak?
Answer:
T₁ = 17°C + 273 = 290K
p₁T₂ = p₂T₁, (From Gay-Lussac’s law)
T₂ = \(\frac{\mathrm{T}_{1} \mathrm{p}_{2}}{\mathrm{p}_{1}}=\frac{290 \times 120}{100}\) = 348K Pi
348 – 273 = 75°C the container might leak.

Question 4.
A cylinder of propane gas at 20°C exerted a pressure of 8.5 atmospheres. When exposed to sunlight it is warmed up to 28°C. What pressure does the gas now experience?
Answer:
T₁ = 20°C = 273 + 20 = 293K
T₂ = 28 + 273 = 301 K
p₂ = P₁ × \(\frac{\mathrm{T}_{2}}{\mathrm{T}_{1}}\) (From Gay-Lussac’s law)
p₂ = \(\frac{8 \cdot 5 a t m \times 301 K}{293 K}\) = 8.73 atm

Question 5.
Calculate the volume of 47.0 cm³ of gas at 27°C, would occupy at 22°C at constant pressure.
Answer:
\(\frac{V_{1}}{V_{2}}=\frac{T_{1}}{T_{2}}\) (from Charle’s law)
V₂ = V₁ × \(\frac{T_{2}}{T_{1}}\)
V₁ = 47.0 cm³, T₁ = 273 + 27 = 300K, T₂ = 273 + 22 = 295K
V₂ = \(\frac{47 \cdot 0 \mathrm{cm}^{3} \times 295 \mathrm{K}}{300 \mathrm{K}}\) = 46.2 cm³

Question 6.
An open flask contains air at 27°C. Calculate the fraction of air that would be expelled out, at 477 °C.
Answer:
Here, gas expands against a constant pressure of 1 atm. Hence Charle’s law can be used.
Now, V₁ = V (say), T₁ = 273 + 27 = 300 K
V₂ = ? T₂ = 273 + 477 = 750 K
∴ V₂ = \(\frac{\mathrm{V}_{1} \mathrm{T}_{2}}{\mathrm{T}_{1}}=\frac{750 \mathrm{V}}{300}\) = 2.5 V
Thus, volume of air expelled = 2.5 V – V = 1.5 V

Question 7.
A 10.0 litre cylinder of H₂ gas is connected to an evacuated 290.0 litre tank. If the final pressure is 700 mm Hg, what must have been the original gas pressure in the cylinder?
Answer:
Initial volume, V₁ = 10.0 litre
Final volume, V₂ = 10.0 + 290.0 = 300.0 litre
Let initial pressure = p₁
Final pressure, p₂ = 700 mm Hg (given)
Applying, p₁V₁ = p₂V₂, we get
p₁ = \(\frac{\mathrm{p}_{2} \mathrm{V}_{2}}{\mathrm{V}_{1}}=\frac{700 \times 300}{10}\) = 21000 mm Hg

Question 8.
Ramu breaths approximately 0.42 L of oxygen at 1 atm with each breath. If an oxygen cylinder holds 8L at 290 atm, how many breaths does the oxygen cylinder supply?
Answer:
P₁v₁ = P₂V₂
1 × V₁ = 290 atm × 8 lit
V₁ = 2320 L
Number of breaths = \(\frac{2320 \mathrm{lit}}{0.42 \mathrm{lit}}\) = 5523.81

Question 9.
A balloon is filled with hydrogen at room temperature. It will burst if pressure exceeds 0.2 bar. If at 1 bar pressure the gas occupies 2.27 L volume, upto what volume can the balloon be expanded?
Answer:
According to Boyle’s law P₁V₁ = p₂V₂
V₂ = \(\frac{p_{1} V_{1}}{p_{2}}=\frac{1 b a r \times 2 \cdot 27 L}{0 \cdot 2 b a r}\) = 11.35L
Since balloon bursts at 0.2 bar pressure, the volume of balloon should be less than 11.35 L.

Question 10.
On a ship sailing in Pacific ocean where temperature is 23.4°C a balloon is filled with 2 L air. What will be the volume of the balloon when the ship reaches Indian ocean, where temperature is 26.1°C?
Answer:
V₁ = 2L
T₂ = 26.1 + 273 = 299.3K,T₂ = (23.4 + 273)K = 296.4K
From Charles’s law
\(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}} \quad \Rightarrow V_{2}=\frac{V_{1} T_{2}}{T_{1}}\)
V₂ = \(\frac{2 \mathrm{L} \times 299 \cdot 1 \mathrm{K}}{296 \cdot 4 \mathrm{K}}\) = 2018 L

Question 11.
At 25°C and 760 mm of Hg 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.
Answer:
p₁ = 760 mm Hg V₁ = 600 ml
T₁ =25 + 273 = 298 K V₂ =640ml
T₂ =10 + 273 = 283 K
According to combined gas law
\(\frac{\mathrm{p}_{1} \mathrm{V}_{1}}{\mathrm{T}_{1}}=\frac{\mathrm{p}_{2} \mathrm{V}_{2}}{\mathrm{T}_{2}}\)
p₂ = \(\frac{p_{1} V_{1} T_{2}}{T_{1} V_{2}}\) = \(\frac{760 \times 600 \times 283}{640 \times 298}\) = 676.6 mm Hg

Question 12.
What is the molar volume of ideal gas at S.T.P.?
Answer:
The volume occupied by one mole of a gas at STP is called molar volume. Molar volume at STP is

• 22.4 L at 273.15 K and 101.325 kpa(l atm)
• 22.71 L at 273.15 K and 100 kpa (1 bar)

Ideal Gas Equation

Question 1.
What is an ideal gas?
Answer:
A gas which obeys Boyle’s law, Charle’s law and Avogadro’s law at all temperatures and pressures is called an ideal gas.

Question 2.
Derive ideal gas equation.
Answer:
According to
Boyle’s Law V α \(\frac{1}{\mathrm{p}}\) (at constant‘T’ and n)
Charle’s Law V α T (at constant ‘p’ and n)
Avogadro’s Law V α n (at constant p and T)
By combining all these laws,
V α \(\frac{\mathrm{nT}}{\mathrm{p}}\)
V = R\(\frac{\mathrm{nT}}{\mathrm{p}}\)
pV = nRT
Where ‘R’ is the proportionality constant. It is same for all gases. Therefore it is also called Universal Gas Constant. This is called ideal gas equation.

Note: This equation is a relation between four variables that describe the state of any gas completely . Therefore, it is also called equation of state.

Question 3.
Calculate the value of R (universal gas constant) in SI units.
Answer:
In SI units, p = 10⁵ N/m² (or Pascal), and the volume is 0.0227098 m³, temperature, T = 273.15K and n = 1 mol. We know that
pV = nRT
R = \(\frac{P V}{n T}\)
R = \(\frac{10^{5} \mathrm{Pa} \times 0 \cdot 02271 \mathrm{m}^{3}}{1 \mathrm{mol} \times 273 \cdot 15 \mathrm{K}}\) = 8.314 Pam³K^-1 mol^-1 = 8.314 JK^-1 mol^-1

Question 4.
Give the value of R in different units.
Answer:
R = 0.0821 L atm /mol/ K
= 8.314 J/mol/K
= 1.987 cal/mol/K
= 62.336 L torr / mol /K
= 0.0831 Lbar/mol/K

Question 5.
Derive combined gas equation from ideal gas equation.
Answer:
The ideal gas equation PV = nRT can be written as
\(\frac{p V}{T}\) = nR = constant for a given mass of gas.
If the pressure, volume and temperature of a given mass of gas is changed p₁, V₁ and T₁ to P₂, V₂, and T₂.
\(\frac{\mathrm{p}_{1} \mathrm{V}_{1}}{\mathrm{T}_{1}}\) = nR and \(\frac{\mathrm{p}_{2} \mathrm{V}_{2}}{\mathrm{T}_{2}}\) = nR
\(\frac{p_{1} V_{1}}{T_{1}}=\frac{p_{2} V_{2}}{T_{2}}\)
This is called combined gas equation.

Question 6.
Derive the relation between density and molar mass of a gas from ideal gas equation.
Answer:
For ideal gas,
PV = nRT
\(\frac{\mathrm{n}}{\mathrm{V}}=\frac{\mathrm{p}}{\mathrm{RT}}\)
But,n = \(\frac{m}{M}\) m = mass of the gas
∴ \(\frac{m}{M V}=\frac{p}{R T}\) M = Molar mass of the gas
But m/V = d
\(\frac{d}{M}=\frac{p}{R T}\) Where d is the density of the gas
On rearranging this equation we get
\(M=\frac{d R T}{p}\)

The partial pressure formula of one gas in a mixture of gases is equal to the amount of pressure that would of the other gases were removed.

Question 7.
State Dalton’s law of partial pressure.
Answer:
It states that “the total pressure exerted by a mixture of non reacting gases is equal to the sum of the partial
pressures of all the component gases at constant temperature”.
It can be expressed as P = p₁ + p₂ + p₃ + …….. + p_n
where P is the total pressure of the mixture of gases and p₁, p₂, p₃ ……. p_n are the partial pressures of component gases.

Question 8.
What is aqueous tension? How do you calculate the pressure of the dry gas from moist gas?
Answer:
Pressure exerted by saturated water vapour is called aqueous tension.
P_{dry gas} = P_tatal – Aqueous tension

Question 9.
What is aqueous tension in dry gas?
Answer:
Zero.

Question 10.
Derive the relation between partial pressure of a gas and Its mole fraction from ideal gas equation.
Answer:
Assuming that each gas behaves ideally, we can calculate the partial pressure of each gas from the ideal gas equation. Let us consider a mixture of three non reacting gases, then, from Dalton’s law, we have

On dividing p₁ by p_total we get

Thus, P₁ = x₁ P_total
Similarly for other two gases, we can write p₂ = x₂ P_total and p₃ = x₃ P_total
Thus a general equation will be:
P_i = x_i P_total
where p_i and x_i are partial pressure and mole fraction of i^th gas respectively. If total pressure of a mixture of gases is known, the equation can be used to find pressure exerted by individual gases.

Numerical Problems

Question 1.
What will be the total pressure when 2g of O₂ and 1.5g of hydrogen are taken in a vessel of volume one litre at 0 °C (R = 0.0821 lit atm mol^-1)
Answer:
pV = nRT

Question 2.
Calculate the total pressure exerted by a mixture of 4.2g of ethane and 6.2g of O₂ contained in 5 dm³ flask at 0°C
Answer:

Question 3.
A neon-dioxygen mixture contains 70.6 g dioxygen and 167.5 g neon. ¡f the pressure of the mixture of gases in the cylinder is 25 bar, what are the partial pressures of dioxygen and neon in the mixture?
Answer:

Question 4.
Calculate the molecular mass of a gas when density is 3.6 kg m^-3 at 273 K and at 1 atm pressure.
Answer:
Molecular mass of the gas = \(\frac{\mathrm{dRT}}{\mathrm{P}}\)

Question 5.
Calculate the temperature of 6.0 mol of a gas occupying 4dm³ at 2 bar. [R = 0.0825 bar dm³ K^-1 mol^-1]
Answer:
Ideal gas equation: PV = nRT

Question 6.
25 cm³ of a gas at 1.01 atm at 25°C was compressed to 15 cm³ at 35°C. Calculate the final pressure of the gas.
Answer:
p₁ = 1.01 atm, p₂ = ?, V₁ = 25 cm³, V₂ = 15 cm³,
T₁ = 25 + 273 = 298 K, T₂ = 35 + 273 = 308 K
\(\frac{\mathrm{p}_{1} \mathrm{V}_{1}}{\mathrm{T}_{1}}=\frac{\mathrm{p}_{2} \mathrm{V}_{2}}{\mathrm{T}_{2}}\)
p₂ = \(\frac{p_{1} V_{1} T_{2}}{T_{1} V_{2}}=\frac{1 \cdot 01 \times 25 \times 308}{15 \times 298}\) = 1.74 atm

Question 7.
How do you calculate the average speed of gas molecules?
Answer:
If there are n number of molecules in a sample and their individual speeds are u₁, u₂,….. u_n, then average speed of molecules uav can be calculated as follows:

Question 8.
Explain Maxwell-Boltzmann distribution speeds of gas molecules.
Answer:
Maxwell-Boltzmann have shown that actual distribution of molecular speeds depends on temperature and molecular mass of a gas.

The following graph shows schematic plot of number of gas molecules vs. molecular speed at two different temperatures T₁ and T₂ (T₂ is higher than T₁). The distribution of speeds shown in the plot is called Maxwell Boltzmann distribution of speeds.

The graph shows that number of molecules possessing very high and very low speed is very small. The maximum in the curve represents speed possessed by maximum number of molecules. This speed is called most probable speed, u_mp. This is very close to the average speed of the molecules.

Question 9.
Define mean square speed (ū²)
Answer:

Question 10.
Define root mean square velocity.
Answer:

Question 11.
Write the relation between u_rms, u_av and u_mp
Answer:
u_mp : u_av : u_rms:: 1 : 1 . 128 : 1 . 224
u_ms > u_av > u_mp

Kinetic Molecular Theory of Gases

Question 1.
Give the postulates of kinetic theory of gases
Answer:
The fundamental postulates of kinetic theory of gases are:

• Gases are made up of extremely small particles called molecules.
• Molecules move randomly in straight lines in all directions at various speeds and the direction of motion is changed when colliding with each other or with walls of the container.
• The volume of a gas molecule is negligible when compared to the total volume occupied by the gas.
• The collisions of gas molecules with the container walls exert the pressure.
• The collisions are perfectly elastic, i.e., no energy loss on collision due to friction.
• There is no intermolecular forces between the gas molecules.
• The average kinetic energy of gas a molecule is directly proportional to the absolute temperature.
  (K.E. ∝ T)

Behaviour of real gases: Deviation from ideal gas behaviour

Question 1.
What are real gases?
Answer:
The gases which do not obey ideal gas equation are called real gases.

Question 2.
Draw the plot of pV vs p for CO, CH₄, H₂, He and ideal gas.
Answer:

Question 3.
Draw the plot of p vs V to show the deviation of real gas from ideal behaviour.
Answer:

Question 4.
Why do gases deviate from the ideal behaviour? What are the causes for the deviations of real gases from ideal behaviour? Or Mention the two conditions of kinetic theory of gases which do not hold good under low temperature and high pressure.
Answer:
The deviation of real gases from ideal behaviour is due to, the two wrong assumptions made in the kinetic theory of gases. They are
(a) There is no force of attraction between the gas molecules.
(b) Volume of a gas molecule is negligible when compared to the total volume occupied by a gas.
If assumption (a) is correct, the gas will never liquify. But gases liquify when cooled and compressed.
The assumption (b) can be true only when volume of the gas is large and pressure is very low. If the pressure is very high, volume of the gas is very low, and hence volume of single gas molecule cannot be neglected.

Question 5.
Under what conditions of temperature and pressure, gases show ideal behaviour?
Answer:
Low pressure and high temperature

Question 6.
What are the conditions under which gases deviate from ideality or ideal behaviour?
Answer:
Under high pressure and low temperatures, real gases show more deviations from ideal behaviour.

Question 7.
Give reason: Gases show ideal behaviour at high temperature and low pressure.
Answer:

• Because no intermolecular forces are operating between the gas molecules at high T.
• Volume of a gas molecule is negligible compared to total volume of gas at low pressure.

Question 8.
Give the difference between ideal and non ideal (real) gases.
Answer:

Ideal gas	Real gas
1. Obeys both Boyle’s law and Charle’s law at all P and T.	1. Does not Obey both Boyle’s law and Charle’s law at all P and T.
2. Obeys ideal gas equation PV = nRT	2. Does not Obey ideal gas equation PV = nRT
3. There are no intermolecular force of attraction between the gas molecules	3. There are weak van der Waal’s force of attraction between the molecules
4. Volume of a gas molecule is negligible when compared to the total volume occupied by the gas	4. Volume of a gas molecule is negligible when compared to the total volume occupied by the gas at high pressures.
5. Cannot be liquefied	5. Can be liquefied.

Question 9.
The pressure exerted by the real gas is lower than the pressure exerted by the ideal gas at high pressures. Explain.
Answer:
At high pressures molecules of real gases are very close to each other. Molecular interactions start opening. At high pressure, molecules do not strike the walls of the container with full impact because these are dragged back by other molecules due to molecular attractive forces. This affects the pressure exerted by molecules on the walls of the container. Thus, the pressure exerted by the gas is lower than the pressure exerted by the ideal gas.

Question 10.
Give the expressions of pressure correction term and volume correction term in real gases.
Answer:
Pressure correction term = P_i = p_real + \(\frac{a n^{2}}{V^{2}}\)
Volume correction term = V_i = V – nb
Where ‘a’ and ‘b’ are van der Waals constants.

Question 11.
Write the van der Waals equation for n moles of gas.
Answer:
\(\left(p+\frac{a n^{2}}{V^{2}}\right)\) (V – nb) = nRT

Question 12.
Write the van der Waals equation for one mole of gas.
Answer:
\(\left(p+\frac{a}{V^{2}}\right)\) (V – b) = RT

Question 13.
What is the importance of van der Waals constants ‘a’ and ‘b’?
Answer:
Value of ‘a’ is measure of magnitude of intermolecular attractive forces with in the gas and ‘b’ measures excluded volume of gas.

Question 14.
Is van der Waals constant ‘a’ dependent on T and P? if no, on what factor it depends.
Answer:
No, it depends on the characteristic of a gas.

Question 15.
When do real gases show ideal behaviour?
Answer:
Real gases show ideal behaviour at low pressure and high temperature.

Question 16.
Name the factor used to measure deviation of real gases from ideal behaviour.
Answer:
Compressibility factor.

Question 17.
Define compressibility factor (Z).
Answer:
It is defined as the ratio of product pV and nRT.
Mathematically, Z = \(\frac{p V}{n R T}\)

Question 18.
What is the value of compressibility factor Z in an ideal gas?
Answer:
For ideal gas, Z = 1

Question 19.
Show the variation of compressibility factor with pressure graphically for the gases N₂, H₂, O₂, CH₄, CO₂ and ideal gas.’
Answer:

Question 20.
Discuss the significance of compressibility factor Z.
Answer:
Case-I: If Z > 1

•  V_real > V_ideal
• The repulsion forces become more significant than the attractive forces.
• The gas cannot be compressed easily.
• Z > 1 for gases like He, and H₂.

Case-II: If Z < 1

• V_real < V_ideal
• The attractive forces are more significant than the repulsive forces.
• The gas can be liquified easily.
• Z < 1 for gases like NH₃, CO₂, SO₂ etc.

Question 21.
The compressibility factor Z for gas “X” is greater than 1. What is the nature of deviation of gas X?
Answer:
Non-ideal gas (real gas) with positive deviation from ideal behaviour.

Question 22.
The compressibility factor Z for gas “X” is less than 1. What is the nature of deviation of gas X?
Answer:
Non-ideal gas (real gas) with negative deviation from ideal behaviour.

Question 23.
The compressibility factor Z for gas “X” is equal to 1. What is the nature of the gas X?
Answer:
Ideal gas.

Question 24.
What is Boyle temperature or Boyle point?
Answer:
The temperature at which a real gas obeys ideal gas laws over an appreciable range of pressure is called Boyle temperature or Boyle point.

Liquifaction of Gases

Question 1.
Name the three critical constants.
Answer:

1. Critical temperature (T_c)
2. Critical pressure (P_c)
3. Critical volume (V_c)

Question 2.
Define critical temperature (T_c)
Answer:
It is the temperature above which a gas cannot be liquified by applying pressure.

Question 3.
Define critical pressure (P_c)
Answer:
It is the minimum pressure required to cause liquifaction of a gas at critical temperature (T_c).

Question 4.
Define critical volume (Vc)
Answer:
It is the volume occupied by one mole of a gas at critical temperature (T_c) and critical pressure (P_c).

Question 5.
Draw the isotherms of CO₂ (Andrews curves) at different temperatures, or Show the relations between V, p and T of CO₂ graphically according to Thomas Andrews curves.
Answer:

Thomas Andrews obtained the critical isotherm of CO₂ and noticed that at high temperatures, isotherms look like that of an ideal gas and the gas cannot be liquified even at very high pressure. The isotherms plotted between P versus V at different temperatures for one mole of CO₂ gas are shown in figure.

Following conclusions can be drawn from these graphs:

At higher temperatures of more than 30.98°C say 50 °C, the isotherms show almost ideal behaviour.

As the temperature is lowered, the isotherms show deviation from ideal behaviour.

At 30.98 °C, carbon dioxide remains as gas up to 73 atm. But liquid appears for the first time at 73 atm (represented by point ‘E’). Hence 5 30.98 °C is called critical temperature for CO₂.

Above 73 atm. there is a steep rise in the pressure. This steep portion of the curve represents the isotherm of liquid state for which small decrease in volume results in steep rise in the pressure.

At even lower temperature, say 20 °C, the liquid appears at point ‘B’. Further compression does not change the pressure up to point ‘C’. After point. C the curve again becomes steep representing the isotherm for liquid CO₂.

Question 6.
What is the critical temperature for CO₂?
Answer:
Tc for CO₂ = 30.98°C

Question 7.
Liquifaction of permanent gas requires both cooling and compression. Why?
Answer:
Compression brings the molecules in close vicinity and cooling slows down the movement of molecules. Therefore, intermolecular interactions may hold the molecules closely and slowly moving molecules together and the gas liquifies.

Question 8.
Critical temperatures of ammonia and carbon dioxide are 405.5 K and 304.10 K respectively. Which of these gases will liquify first when you start cooling from 500 K to their critical temperature?
Answer:
Ammonia will liquify first because its critical temperature will be reached first. Liquifaction of C02 will require more cooling.

Question 9.
Justify the statement “Lower the critical temperature easier is the liquifaction of the gas”
Answer:
If critical temperature is low, intermolecular forces are more. So, the gas can be liquefied easily. At higher critical temperatures, gases behave ideally and hence no intermolecular forces. Hence they cannot be liquified easily.

Liquid State

Question 1.
Give any three characteristics of liquid state.
Answer:

1. A liquid has definite volume but no fixed shape. They acquire the shape of the container.
2. Liquids are also incompressible.
3. In liquids, particles move randomly but slowly.
4. The particles do rotate and vibrate. The degree of freedom is more in case of liquids compared to solids.

Question 2.
What is equilibrium vapour pressure or saturated vapour pressure?
Answer:
The saturated vapour pressure of a liquid is the pressure exerted by its vapour when the liquid and vapour are in dynamic equilibrium.

Question 3.
Define boiling point.
Answer:
The temperature at which the vapour pressure is equal to the atmospheric pressure is called the boiling point. If atmospheric pressure is equal to 1 atmosphere, the boiling point is called normal boiling point.

Question 4.
At what pressure, the boiling point of the liquid is taken as normal boiling point.
Answer:
1 atm

Question 5.
At what pressure, the boiling point of the liquid is taken as standard boiling point.
Answer:
1 bar

Question 6.
Standard boiling point of the liquid is slightly lower than the normal boiling point. Why?
Answer:
Because 1 bar pressure is slightly less than 1 atm pressure.

Question 7.
What is the normal boiling point of water?
Answer:
100°C (373 K)

Question 8.
What is the standard boiling point of water?
Answer:
99.6°C (372.6 K)

Question 9.
Liquids at high altitudes boil at lower temperatures in comparison to that at sea level. Why?
Answer:
At high altitudes, atmospheric pressure is low.

Question 10
Pressure cooker is used for cooking food on hills. Why?
Answer:
Because water boils at low temperature on hills.

Question 11.
In hospitals, surgical instruments are sterilised in autoclaves only. Why?
Answer:
Because boiling point of water is increased by increasing pressure above the atmospheric pressure by using a weight covering the vent.
Note: Surface tension of liquid is due to the unbalanced force experienced by molecules at the surface of a liquid.

Question 12.
Define surface tension.
Answer:
Surface tension is defined as, the force acting per unit length perpendicular to the imaginary line drawn on the surface of liquid.

Question 13.
What is the dimensional unit of surface tension?
Answer:
kg s^-2

Question 14.
What is the SI unit of surface tension?
Answer:
N m^-1.

Question 15.
What is surface energy?
Answer:
The energy required to increase the surface area of the liquid by one unit is called surface energy.

Question 16.
On which factor the magnitude of surface tension depends?
Answer:
On the attractive forces between the molecules. Student’s illuminator

Question 17.
In what way the surface tension of the liquid depends on attractive forces?
Answer:
Larger the attractive forces, larger the surface tension of the liquid.

Question 18.
What is the effect of temperature on surface tension of the liquid?
Answer:
As the temperature increases, surface tension of liquid decreases due to decrease in intermolecular attraction.

Question 19.
Define viscosity of a liquid.
Answer:
Viscosity of a liquid is a measure of resistance to flow, which arises due to the internal friction between layers of fluid as they slip one another while liquid flows.
Note: Strong intermolecular forces between molecules held them together and resist movement of layers one another.

Question 20.
What is laminar flow?
Answer:
The type of flow in which there is a regular gradation of velocity in passing from one layer to the next is called laminar flow.

Question 21.
Give the expression to calculate the force required to maintain the flow of layers.
Answer:
F = nA\(\frac{d u}{d z}\)
where n is proportionality constant called co-efficient of viscosity. A is the area of cross section, \(\frac{d u}{d z}\) is the velocity gradient.

Question 22.
Define co-efficient of viscosity. Give its SI unit.
Answer:
Co-efficient of viscosity is the force when velocity gradient is unity and the area of contact is unit area.
SI unit of co-efficient of viscosity = Nsm^-2 or Pa.s or kg m^-1 s^-1.
Note:

• In CGS system, the unit of co-efficient of viscosity is poise.
  1 poise = 1 g cm^-1 s^-1 = 10^-1 kg m^-1 s^-1
• Hydrogen bond and van der wall forces are strong enough to cause high viscosity.

Question 23.
Viscosity of liquids decreases with increase in temperature. Why?
Answer:
Because at high temperature molecules have high kinetic energy and can over come intramolecular forces to slip one another between the layers.