Solutions-Class-11-Science-Chemistry-Chapter-10-States of Matter-Maharashtra Board

(d) poise

(c) PV = 2RT

(c) both (a) and (b)

(a) nil

(b) 11.2 dm³

Different gases can be mixed by diffusion.

The pressure that each individual gas would exert if it were alone in the container containing gaseous mixture is called partial pressure.

When the gas heated at constant pressure the volume of the gas increases.

As a bubble of methane gas rises from the bottom of the sea, the pressure above the

bubble due to water level decreases hence the size of the bubble increases.

(a) T = 273 - 15 = 258 K

(b) T = 273 + 25 = 298 K

(c) T = 273 - 197 = 76 K

(d) T = 273 + 273 = 546 K.

(a) Pressure = 10 x 101325 Pa = 1013250 Pa

(b) Pressure = 1 kPa = 1 x 10³ Pa

(c) Pressure = 107000 Nm^-2 = 107000 Pa

(d) Pressure = 1 atmosphere = 101325 Pa.

(a) Pressure = 1.5 x 101325 Pa = 151987.5 Pa

(b) Pressure = 89 kPa = 89 x 10³ Pa = 89 x 10³ Nm^-2

(c) Pressure = 101.325 kPa = 101.325 x 10³ Pa

= \(\frac{101.325×10^3}{1×10^5}\) bar

= 101.325 × 10^-2 bar

= 1.01325 bar

(d) Temperature = 273 + (-100) = 173 K

(e) Pressure = 0.124 torr = \(\frac{0.124}{760}\) atm = 1.631 x 10^-4 atm.

(a) Correct          (b) Correct          (c) Correct.         (d) Correct.

The reaction represented in (a) is in accordance with the principle of stoichiometry.

The air inside the balloons has low density due to low pressure and high volume. Hence the air balloons float in air. This can be explained by Boyle's law.

(a) Boyle's law

(b) Charles law

(c) Avogadro law.

(a) At constant temperature since the pressure is doubled, the volume will become half.

(b) At constant pressure temperature is decreased 1 times i.e. 300 K to 150 K. Hence, the volume of the gas will become half, as shown in (a).

(c) Let the initial volume of the gas be V.

P₁ = 4 bar, P₂ = 3 bar, T₁ = 400 K, T₂ = 300 K

If V' is the final volume then,

\(\frac{P_1V}{T_1}=\frac{P_2V'}{T_2}\)

∴ \(\frac{V'}{V}=\frac{P_1×T_2}{P_2×T_1}\) = \(\frac{V'}{V}=\frac{4×300}{3×400}\) = 1

∴ V’ = V

Since the volume of the gas remains the same the piston will remain at the same position.

The gaseous state is characterized by the following physical properties :

• Gases are lighter than solids and liquids, resulting in lower density.
• Gases occupies entire space and takes container shape.
• Gas molecules are widely separated and in continuous motion, exerting equal pressure.
• Intermolecular forces are weakest in gases.
• Gases possesses diffusion property, a spontaneous homogenous intermixing of gases.
• Gases are highly compressible.

The distortion in the electron cloud caused by the attractive and repulsive interactions between the nuclei of two approaching non-polar molecules is known as polarizability.

• More electrons in a molecule increase polarisability.
• Defines how external electric field can alter an atom's or molecule's electron cloud.

A hydrogen bond is a special type of dipole-dipole attraction which occurs when a hydrogen atom is bonded to a strongly electronegative atom or an atom with a lone pair of electrons.

Aqueous tension : In case of pure water or an aqueous solution, there exists saturated water vapour in equilibrium with it on its surface and the pressure of this vapour is called aqueous tension.

Dipole moment (μ) of al polar molecule is the product of the magnitude of the charge (Q) and the distance (r) between the centres of positive and negative charges. It is designated by a Greek Letter (u) and its unit is Debye (D).

It is represented as, μ = Q × r

• Dipole moment is a vector quantity and represented by a small arrow with tail at the positive centre and head towards the negative centre.
• The polar molecules have permanent dipole moment.

Atmospheric pressure is crucial for the process of drinking water through a straw. It reduces the pressure inside the straw, allowing the liquid to be pushed up to the mouth. However, the pressure difference on the top of Mount Everest is very low, making it difficult to drink water with a straw compared to the base.

(a) Hydrogen bonding

(b) London dispersion forces

(c) Dipole-dipole interactions

(d) Dipole-dipole interactions.

(a) London dispersion forces

(b) London dispersion forces

(c) Dipole-dipole interactions

(d) Hydrogen bonding.

(a) Boyle's law-at constant temperature

(b) Charles' law-at constant pressure.

(a) Boyle’s law : For a fixed mass (number of moles ‘n’) of a gas at constant temperature, the pressure (P) of a gas is inversely proportional to the volume (V) of gas.

(b) Charles’ law  : It states that at constant pressure, the volume of the given mass of a gas is directly proportional to its absolute temperature.

• It states that at constant pressure the volume of a given mass of a gas increases or decreases by \(\frac{1}{273.15}\) of its volume at 0 °C for every degree rise or fall of temperature.

Statement : The total pressure of a mixture of two or more unreactive gases is the sum of the partial pressures of the individual gases in the mixture.

Example : If P₁, P₂, P₃, .... .. are partial pressures of constituent gases in the mixture at constant volume and temperature then by Dalton’s law, total pressure is,

P_Total = P₁ + P₂ + P₃ + .... ..

This is mathematical statement of Dalton’s law.

Derivation of Ideal Gas Equation :

The three gas laws, namely, Boyle’s law, Charles law and Avogadro law, are combined mathematically to obtained what is called ideal gas equation.

Consider n moles of a gas occupying volume V at pressure P and temperature T.

(i) By Boyle's law, at constant temperature,

V ∝ 1/P (at constant n and T)

(ii) By Charles’ law, at constant pressure,

V ∝ T (at constant n and P)

(iii) By Avogadrds law, at constant pressure and temperature,

V ∝ n (at constant P and T)

Hence,

V ∝ \(\frac{n×T}{P}\)

∴ V = \(\frac{nRT}{P}\)    OR   PV = nRT

where R is a proportionality constant and called gas constant or universal gas constant.

The equation, PV = nRT is called an ideal gas equation or an equation of state.

• If m is the mass of the gas then, number of moles, n = m/M Hence, ideal gas equation can be written as, PV = \(\frac{mRT}{M}\).

Combined Gas law :

Consider n moles of a gas. By ideal gas equation,

PV = nRT

\(\frac{PV}{T}\) = nR = constant. (Since n and R are constant)

Hence, for a given gas,

\(\frac{P_1V_1}{T_1}\) = \(\frac{P_2V_2}{T_2}\) 

This equation is also called Combined gas law or combined gas equation.

(a) P ∝ \(\frac{1}{V}\) 

(b) Boyle's law

(c) Statement : For a fixed mass (number of moles ‘n’) of a gas at constant temperature, the pressure (P) of a gas is inversely proportional to the volume (V) of gas.

• Gases consist of tiny particles (molecules or atoms).
• On average, gas molecules remain far apart from each other. Therefore, the actual volume of the gas molecules is negligible as compared to the volume of the container. Hence, gases are highly compressible.
• The attractive forces between the gas molecules are negligible at ordinary temperature and pressure. As a result, the gas expands to occupy the entire volume of the container.
• Gas molecules are in constant random motion and move in all possible directions in straight lines. They collide with each other and with the walls of the container.
• Pressure of the gas is due to the collision of gas molecules with the walls of the container.
• The collisions of the gas molecules are perfectly elastic in nature, which means that the total energy of the gaseous particle remains unchanged after the collision.
• The different gas molecules move with different velocities at any instant and hence have different kinetic energies.
• However, the average kinetic energy of the gas molecules is directly proportional to the absolute temperature.

The pressure exerted by the vapour in equilibrium with the liquid is known as saturated vapour pressure or simply vapour pressure.

The vapour pressure of water is also called Aqueous Tension.

Explanation :

• All liquids have a tendency of evaporation, during which liquid molecules escape into a gaseous state (vapour).
• If the vessel is closed then the vapour molecules get accumulated on the liquid surface.
• These molecules due to random motion collide with each other and on the walls of a container and get condensed back into the liquid.
• When an equilibrium is attained between evaporation and condensation the number of vapour molecules remain constant giving a definite vapour pressure at constant temperature.
• Mercury manometer may be used to measure the vapour pressure of a‘ liquid as shown in below Fig.

Surface Tension :

The force acting per unit length perpendicular to the line drawn on the surface of liquid is called surface tension.

• Unit : Surface tension is measured in SI Unit, Nm^-1, denoted by Greek letter “γ’’ (Gamma)

Explanation:

• In the bulk of a liquid, every molecule is surrounded equally and symmetrically by other neighbouring molecules of the liquid. Hence, all bulk molecules experience balanced forces.
• But on the surface, the molecules experience downward attractive forces from the molecules in the bulk.
• Since there are no molecules above the liquid surface they experience net force of attraction in the downward direction towards the bulk of the liquid. Therefore, the liquid surface behaves like a ‘stretched membrane which is always under a tension. This gives rise to a surface tension on the liquid surface.

Application of surface tension :

• Cleaning action of soap and detergent is due to the lowering of interfacial tension between water and oily substances. Due to lower surface tension, the soap solution penetrates into the fibre, surrounds the oily substance and washes it away.
• The effective use of toothpastes, mouth washes and nasal drops is due to the presence of substances having lower surface tension. This increases the efficiency of their penetrating action.

Viscosity is the force of friction between the successive layers of a flowing liquid.

OR It is the resistance to the flow of a liquid.

Explanation :

• When a liquid flows, it is considered to be moving in different layers in contact with one another.
• There is a force of friction between the adjacent layers of a moving liquid which opposes the motion of the liquid.
• When a liquid flows over a fixed surface, the layer of the molecules in the immediate contact of the surface experiences maximum friction and it is considered stationary.
• As the distance of upper layers increases from the fixed layer the velocity increases. Hence, there is a regular gradation of velocity of moving liquid layers from one layer to the next, and this flow is called laminar flow.

T = 273 + 25 = 298 K

P₁ = 1 bar; V₁ = 2.27 L, P₂ = 0.3 bar; V₂ = ?

P₁ × V₁ = P₂ × V₂

∴ V₂ = \(\frac{P_1V_1}{P_2}\) = \(\frac{1×2.27}{0.3}\) = 7.567 L

Limit of volume of balloon that can stay inflated = 7.567 L

V₁ = 10 cm³; P₁ = 1 atm, P₂ = 3.5 atm; V₂ =?

P₁ × V₁ = P₂ × V₂

∴ V₂ = \(\frac{P_1V_1}{P_2}\) = \(\frac{1×10}{3.5}\) = 2.857 atm

Final pressure = 2.857 atm.

(a) V₁ = 2 dm³; T₁ = 273 + 0 = 273 K, T₂ = 273 + 10 = 283 K; V₂ =?

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

∴ V₂ = \(\frac{V_1T_2}{T_1}\) = \(\frac{2×283}{273}\) = 2.073 dm³

Volume of a gas = 2.073 dm³.

(b) V₁ = 2 dm³; T₁ = 273 K, T₂ = 273 - 10 = 263 K; V₂ =?

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

∴ V₂ = \(\frac{V_1T_2}{T_1}\) = \(\frac{2×263}{273}\) = 1.927 dm³

Volume of a gas = 1.927 dm³.

V₁ = 2800 m³; T₁ = 273 + 99 = 372 K, T₂ = 273 + 80 = 353 K; V₂ =?

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

∴ V₂ = \(\frac{V_1T_2}{T_1}\) = \(\frac{2800×353}{372}\) = 2657 m³

V₁ = 22.4 L; T₁ = 273 + 0 = 273 K, V₂ = 25 L T₂ =?

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

∴ T₂ = \(\frac{V_2T_1}{V_1}\) = \(\frac{25×273}{22.4}\) = 304.7 K

∴ t₂ = 304.7 - 273 = 31.7 C

T₂ = 304.7 K; t₂ = 31.7 C.

Initial conditions:

Volume V₁ = 20 L, Temperature T₁ = 37 ⁰C = 273 + 37 = 310  K

Moles of He gas n₁ = 0.650 mol, Pressure P₁ = 628.3 bar

Final conditions:

Volume V₂ = 12 L, Temperature T₂ = 177 ⁰C = 273 + 177 = 450 K

Total moles of He gas n₂ = 0.650 mol + 1.25 mol = 1.9 mol

Using the ideal gas law PV=nRT, we can write:

P₁V₁ = n₁RT₁  and P₂V₂ = n₂RT₂

Since R (the gas constant) is the same, we get:

\(\frac{P_1V_1}{n_1T_1}\) = \(\frac{P_2V_2}{n_2T_2}\)

Now, solving for the final pressure P₂:

P₂ = \(\frac{P_1V_1n_2T_2}{n_1T_1V_2}\)

Substituting the values:

P₂ = \(\frac{628.3×20×1.9×450}{0.650×310×12}\)

P₂ ≈ 4443 bar = 4443 x 0.987 atm = 4385 atm

The new pressure inside the container would be approximately 4443 bar = 4385 atm. 

V = 2.31 L; T = 273 + 32 = 305 K, P = 4.7 atm; n = ?

PV = nRT

∴ n = \(\frac{PV}{RT}\) = \(\frac{4.7×2.31}{0.0821×305}\) = 0.4354 mol

Number of moles of the gas = 0.4354 mol.

P₁ = 760 mm, V₁ = 600 mL, T₁ = 273 + 25 = 298 K

V₂ = 640 mL; T₂ = 273 + 10 = 283 K; P₂ = ?

\(\frac{P_1V_1}{T_1}\) = \(\frac{P_2V_2}{T_2}\)

∴ P₂ = \(\frac{P_1V_1T_2}{T_1V_2}\) = \(\frac{760×600×283}{298×640}\) = 676.6 mm Hg

Pressure of gas = 676.6 mm Hg.

m_O2 = 70.6 g, M_O2 = 32 g mol^-1;

m_Ne = 167.5g, M_Ne = 20.18 g mol^-1

P_Total = 25 bar

P_O2 = ? P_Ne =?

n_O2 = \(\frac{m_{O_2}}{M_{O_2}}\) = \(\frac{70.6}{32}\) = 2.206 mol

n_Ne = \(\frac{m_{Ne}}{M_{Ne}}\) = \(\frac{167.5}{20.18}\) = 8.3 mol

Total number of moles = 2.206 + 8.3 = 10.506 mol

Mole fractions :

x_O2 = \(\frac{2.206}{10.506}\) = 0.2099

x_Ne = \(\frac{8.3}{10.506}\)  = 0.79

Partial pressures :

P_O2 = x_O2 x P_Total = 0.2099 x 25 = 5.247 bar

P_Ne = x_Ne x P_Total = 0.79 x 25 = 19.75 bar

Partial pressure of O₂(g) = 5.247 bar.

Partial pressure of Ne(g) = 19.75 bar.

n = 1 mol; V = 2.0 dm³; T = 273 + 20 = 293 K P = ?

PV = nRT

∴ P = \(\frac{nRT}{V}\) = \(\frac{1×0.0821×293}{2.0}\) = 12.02 atm

Pressure = 12.02 atm.

n = 1 mol; T = 273 + 20 = 293 K, P = 101.35 kPa = 101.35 x 10³ Pa,

V =?

PV = nRT

V = \(\frac{nRT}{P}\) = \(\frac{1×8.314×293}{101.35}\) = 24.035 × 10^-3 m³

= 24.035 dm³

Volume of the gas = 24.035 dm³.

P = 2 x 10² kPa = 2 x 10⁵ Pa; V = 0.50 m³ , T = 300 K; n = ?

Number of molecules = ?

PV = nRT

∴ n_CH4 = \(\frac{PV}{RT}\) = \(\frac{2×10^2×0.50}{8.314×300}\) = 40 mol

∴ Number of molecules = n_CH4 x N_A

= 40 × 6.02 × 10²³

= 2.408 × 10²⁵

Number of molecules of CH₄ = 2.408 x 10²⁵