States Of Matter

Given that P₁ = 1 bar.

V₁ = 500 dm³

V₂ = 200dm³

As temperature is constant, the final pressure can be calculated by Boyle's law,

P₁V₁ = P₂V₂

⇒

⇒ P₂ =

⇒ P₂ = 2.5 bar

Therefore, minimum pressure required is 2.5 bar.

According to Boyle’s Law P₁V₁= P₂V₂

Here the temperature is constant. Therefore

1.2 X 120 = P₂X 180

OR

P₂= 1.2 X 120 /180 = 0.8 bar

The equation of state is given by,

pV = nRT ……….. (i)

Where,

p-Pressure of gas

V- Volume of gas

n- Number of moles of gas

R- Gas constant

T -Temperature of gas

From equation (i) we have,

Replacing n = m/M we have

⇒

because the number of moles is given as:

Where,

M - Mass of gas

M - Molar mass of gas

But,

(where, d = density of gas)

Thus, from equation (ii), we have

⇒

⇒

Molar mass (M) of a gas is always constant and therefore, at constant temperature T,

Hence, at a given temperature, the density (d) of gas is proportional to its pressure (p).

Density (d) of the substance at temperature (T) can be given by the expression,

⇒

Now, density of oxide (d₁) is given by,

⇒

Where, M₁ and P₁ are mass and pressure of the oxide respectively.

Density of dinitrogen gas (d₂) is given by,

⇒ d₂ =

Where,M₂ and P₂ are the mass and pressure of the oxide respectively.

According to the given question,

⇒ d₁ = d₂

Molecular mass of nitrogen, M₂ = 28 g/mol

⇒ M₁P₁ = M₂P₂

Therefore, we can write

Hence molar mass

⇒

⇒ M = 70 g/mol

Hence, the molecular mass of the oxide is 70 g/mol

For ideal gas A, the ideal gas equation is given by,

⇒ P_aV = n_aRT

Where, Pa and n_a represent the pressure and number of moles of gas A.

For ideal gas B, the ideal gas equation is given by,

⇒ P_bV = n_bRT

Where,P_a and n_b represent the pressure and number of moles of gas B.

[V and T are constants for gases A and B]

Therefore, we have,

⇒

Where, M_a and M_b are the molecular masses of gases A and B respectively.

Now, we have

Substituting the values

m_a = 1 g

P_a = 2 bar

m_b = 2 g

P_b = 1 bar.

Given,

(Since total pressure is 3 bar)

Substituting these values in equation we have

⇒

Thus, a relationship between the molecular masses of A and B is given by

⇒ 4M_a = M_b

The reaction of aluminium with caustic soda can be represented as:

At STP (273.15 K and 1 atm), 54 g (2 × 27 g) of Al gives 3 × 22400 ml of Hydrogen

⇒ 0.15 g Al gives i.e. 3×22400×0.15/54 = 186.67 ml of Hydrogen.

At STP,

Let the volume of dihydrogen V₂ be at p₂ = 0.987 atm (since 1 bar = 0.987 atm) and T₂ = 293.15K.

⇒ 20°C = (273.15 + 20) K = 293.15 K.

⇒ = constant

So,

⇒

⇒ V₂ = 203ml

Therefore, 203 ml of dihydrogen will be released.

Let the partial pressure of hydrogen in the vessel be P_a.

P₁ = 0.8 bar,

V₁ = 0.5L

V₂ = 1L

P₂ = P_a

Now, at constant temperature

⇒ P₁V₁ = P₂V₂

⇒

P_a = 0.4 bar

Similarly, for oxygen we get the partial pressures of oxygen as

⇒

⇒ P_b = 1.4 bar

It is known that,

The total pressure exerted is the sum of partial pressures so total pressure is (1.4+0.4) = 1.8 bar.

For an ideal gas

Density ρ = p x M/ RT

F rom the given data,

5.46 = 2 x M/ 300 x R ………(1)

ALSO ρ_STP = 1 x M/ 273 x R ………..(2)

FROM (1) & (2) ,WE GET

ρ_{STP =} (1 x M/ 273 x R) x (300 x R/ 2 x M)

= 300 / 273 X 2 = 3.00 g/dm³

Given,

p = 0.1 bar

V = 34.05 ml = 34.05 × 0.001 L = 34.05 × 0.001dm³

R = 0.083 bar dm³/K mol

T = 546°C = (546 + 273) K = 819 K

The number of moles (n) can be calculated using the ideal gas equation as:

PV = nRT

Where, P - Pressure of gas

V - Volume of gas

n - Number of moles of gas

R- Gas constant

T -Temperature of gas

⇒

n = 5.01×10^-5 mol

Therefore, molar mass of phosphorus = mass/moles

⇒

⇒ M = 1247.5 g/mol

Hence, the molar mass of phosphorus is 1247.5 g/mol.

Given:

Initial temperature, T₁ =27°C = (27 + 273) K = 300K

Final temperature, T₂ = 477°C (477 + 273) K = 750K

Let the volume of the round bottomed flask be V.

Then, the volume of air inside the flask at 300K is, V₁ = V.

Let the volume of air inside the flask at 750K is = V₂

According to Charlee’s law we know that: -

(∵ V₁ = V)

∴ Volume of air expelled out = 2.5 V – V = 1.5 V

Hence, the fraction of the air expelled out is given by: -

∴ The fraction of the air expelled out is: 3/5.

Given:

Number of moles, n = 4

Volume, V = 5 dm³

Pressure, P = 3.32 bar

Universal gas constant, R = 0.083 bar dm³ K^–1 mol^–1

Let the temperature be T.

Then, from ideal gas equation we know that: -

PV = nRT

= 50K

Hence, the temperature of 4.0 mol of a gas occupying 5 dm³ at 3.32 bar is 50K.

Given:

Mass of dinitrogen = 1.4 g

We know that molar mass of dinitrogen = 28 g mol^-1

∴ The number of moles of dinitrogen in 1.4g = 1.4/28 = 0.05mol

Now, the number of molecules in 1.4g of dinitrogen: -

= 0.05 x 6.022 × 10²³

= 3.01 x 10²³

We know that 1 molecule of dinitrogen contains 14 electrons.

∴ 3.01 × 10²³ molecules of dinitrogen contain:

= 14 × 3.01 × 1023

= 4.214 × 10²³ electrons.

Hence, the total number of electrons present in 1.4 g of dinitrogen gas is 4.214 × 10²³.

We know that Avogadro number = 6.02 × 10²³.

Given that rate of distribution = 10¹⁰ grains per second.

Thus, time required to distribute one Avogadro number of wheat grains: -

= Avogadro number / rate of distribution

= 6.023 x 10¹³ s

=1.909 x 10⁶ years

Hence, the time taken to distribute one Avogadro number of wheat grains is 1.909 x 10⁶ years.

Given:

Mass of dioxygen= 8 g

Mass of dihydrogen= 4 g

Now, molar mass of dioxygen = 32g

Thus, number of moles of dioxygen = 8/32 = 0.25 mole

Also, molar mass of dihydrogen = 2g

Thus, number of moles of dihydrogen = 4/2 = 2 mole

Therefore, total number of moles in the mixture:

= 0.25 + 2 = 2.25 mole

Given: -

V = volume = 1 dm³,

n = number of moles = 2.25 mol

R = universal gas constant = 0.083 bar dm³ K^-1 mol^-1

T = temperature = 27°C = 300 K

Let the total pressure be P.

From ideal gas equation we know that -

PV = nRT

= 56.025 bar.

Hence, total pressure of the mixture is 56.025 bar.

Given

Radius of the balloon, r = 10 m

∴ Volume of the balloon = 4/3 x πr³

= 4/3 x 22/7 x 10³

= 4190.5 m³

Thus, the volume of the displaced air is 4190.5 m³.

Also, given that

Density of air = 1.2 kg m^-3

Then, mass of displaced air = 4190.5 × 1.2 kg

= 5028.6 kg

From ideal gas equation we know that: -

PV = nRT

⇒ PV = m/M x RT

Where, P = pressure = 1.66bar

V= volume = 4190.5 m³

R = universal gas constant = 0.083 bar dm³ K^–1 mol^–1

T= temperature = 27 + 273 = 300K

m = mass of helium inside the balloon

M = molar mass of helium = g

∴

=1117.5kg

Now, total mass of the balloon filled with helium:

= (100 +1117.5) kg

= 1217.5 kg

Hence, pay load = (5028.6 – 1217.5) kg = 3811.1 kg

∴The pay load of the balloon is 3811.1 kg.

From ideal gas equation we know that: -

PV = nRT

⇒ PV = m/M x RT

Where, P = pressure = 1 bar

R = universal gas constant = 0.083 bar dm³ K^–1 mol^–1

T= temperature = 31.1 +273 = 304.1 K

m = mass of CO₂ = 8.8 g

M = molar mass CO₂ = 44g

Let the volume be V

∴

= 5.05L

∴ The volume occupied is 5.05 L.

From ideal gas equation we know that: -

PV = nRT

⇒ PV = m/M x RT

∴

Where, P = pressure of dihydrogen

R = Universal gas constant

T= temperature of dihydrogen

m = mass of dihydrogen

M = molar mass of dihydrogen

V = volume of dihydrogen

∴

Let M be the molar mass of the unknown gas, pressure remains constant. Volume (V) occupied by the unknown gas can be calculated in the same way as above:

∴

According to the question,

∴

= 40gmol^-1

Hence, the molar mass of the gas is 40 g mol^-1.

Let the weight of dihydrogen be 20 g and the weight of dioxygen be 80 g.

Then, the number of moles of dihydrogen:

n(H₂) = 20/2 = 10 moles

The number of moles of dioxygen:

n(O₂) = 80/32 = 2.5 moles

Given that,

Total pressure of the mixture, p(total) = 1 bar

Then, partial pressure of dihydrogen:

= 0.8 bar

Hence, the partial pressure of dihydrogen is 0.8 bar.

We know that

The SI unit for pressure, p is Nm^-2.

The SI unit for volume, V is m³.

The SI unit for temperature, T is K.

The SI unit for the number of moles, n is mol.

Therefore, the SI unit for quantity pV²T²/n is: -

= Nm⁴K²mol^-1.

Charles’ law states that at constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature.

From the above graph between V and T we can find that at constant pressure if temperature is lowered, there is also a decrease in the volume at constant pressure. The volume should therefore become zero at –273°C. Any further lowering is impossible as it will imply negative volume. Hence, –273°C is the lowest possible temperature.

The Critical temperature of any gas gives us an idea regarding the intermolecular forces of attraction between the molecules of that gas. Higher the intermolecular forces of attraction, higher will be the critical temperature. Hence, intermolecular forces of attraction are stronger in the case of CO₂ as critical temperature for carbon dioxide is comparatively higher.

Physical significance of ‘a’:

‘a’ is a measure of the magnitude of intermolecular attractive forces within a gas. This implies that if a molecule has a large value of a then the intermolecular forces of attraction between its molecules will be large and hence it could be easily liquefied.

Physical significance of ‘b’:

‘b’ is a measure of the volume of a gas molecule. It gives us the volume occupied by all the gaseous molecules contained in a given vessel which was earlier neglected according to kinetic theory and this correction term becomes more significant at high pressure as the molecules are themselves incompressible and occupy an appreciable fraction of the total volume.