Chapter 5 States of Matter

Volume of gas(V₁) = 500dm³   (given)

Pressure of gas (p₁) = 1 bar    (given)

Volume of compressed gas (V₂) = 200dm³  (given)

Now ,let P₂ be the pressure required to compress the gas.

Therefore at constant temperature P₁ V₁ = P₂ V₂ (Boyle’s Law)

As a result P₂ = P₁ V₁ / V₂ = 1 X 500/200 = 2.5 bar

Therefore, the minimum pressure required is 2.5 bar.

According to Boyle’s Law P₁ V₁ = P₂ V₂

Here the temperature is constant. Therefore

1.2 X 120 = P₂ X 180

OR

P₂ = 1.2 X 120 /180 = 0.8 bar

Therefore, the pressure would be 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,

p = n RT/V

Where n= Mass of gas(m)/ Molar mass of gas(M)

Putting value of n in the equation, we have

 p = m RT/ MV ------------(ii)

Now density(ρ) = m /V ----------------(iii)

Putting (iii) in (ii) we get

P = ρ RT / M

OR

ρ = PM / RT

Hence, at a given temperature, the density (ρ) of gas is proportional to its pressure (P)

Using the relationship of density (ρ) of the substance at temperature (T)

We have ρ = Mp/RT

or p = ρ RT/ M

For the given data if M is the molar mass of the gaseous oxide, we have

2 = ρ RT/ M ……………(1)

Also for nitrogen 5 = ρ RT/28 ………………….(2)

From (1) & (2), we have

5/2 = M/28

Or

M = 5 X 28/ 2 = 70g/mol

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

Mass of gas A , W_A = 1g

Mass of gas B,  W_B = 2g

Pressure exerted by the gas A = 2 bar

Total pressure due to both the gases = 3 bar

In this case temperature & volume remain constant

Now if M_A & M_B are molar masses of the gases A & B respectively,therefore

p_A V= W_A RT/M_A & P_total V = (W_A/M_A + W_B/ M_B) RT

= 2 X V = 1 X RT/M_A & 3 X V = (1/M_A + 2/M_B)RT

From these two equations, we get

3/2 = (1/M_A + 2/M_B) / (1 /M_A) = (M_B + 2M_A) / M_B

This result in 2M_A/ M_B = (3/2) -1 = ½

OR M_B = 4M_A

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

4M_A = M_B   

Aluminium reacts with caustic soda in accordance with the reaction

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

Therefore volume of hydrogen at STP released when 0.15g of Al reactsv

=0.15 x 3 x 22.4 /54 = 187ml

Now P₁ = 1 bar,

P₂ = 1 bar

T₁ = 273 K

T₂ = 20 + 273 = 293 K

V₁ = 187 ml

V₂ = x

When pressure is held constant,then

V₂ = P₁ V₁ T₂ / P₂ T₁

OR

x = 1 X 187 X 293 / 0.987 X 273

= 201 ml

Therefore, 201 mL of dihydrogen will be released.

Given,

Mass of carbon dioxide = 4.4 g

Molar mass of carbon dioxide= 44g/mol

Mass of methane  = 3.2g

Molar mass of methane = 16g/mol

Now amount of methane, n_CH4 = 3.2/ 16 = 0.2 mol

& amount of CO₂  n_CO2  = 4.4/ 44 = 0.1 mol

Also we know ,

Pv = (n_CH4 + n_CO2) RT

OR P X 9 = (0.2 +0.1) x 0.0821 x 300

Or  p = 0.3 x 0.0821 x 300 / 9 = 0.821 atm

Hence, the total pressure exerted by the mixture is 0.821 atm

From the equation Pv = n RT for the two gases. We can write

0.8 x 0.5 = n_H2 x RT  or n_H2  = 0.8 x 0.5 / RT

ALSO  0.7 x 2.0  = n₀₂ . RT  or n₀₂  = 0.7 x 2 / RT

When introduced in 1 L vessel, then

Px1L = (n₀₂ + n_H2)  RT

Putting the values, we get

P = 0.4 + 1.4 = 1.8 bar

Hence, the total pressure of the gaseous mixture in the vessel is 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) x 5.46

= 300 x 5.46 / 273 X 2 = 3.00 g/dm³

Hence, the density of the gas at STP will be 3 g dm^–3.

Given,

p = 0.1 bar

V = 34.05 mL = 34.05 × 10^–3 L = 34.05 × 10^–3 dm³

R = 0.083 bar dm³ K^–1 mol^–1

T = 546°C = (546 + 273) K = 819 K

From the gas equation PV = w. RT / M, we get

M = w. RT/ Pv    ……….(1)

Substituting the given values in the equation (1), we get

M = (0.0625 / 0.1 x 34.04) X 82.1 X 819 = 123.46 g/mol

Hence, the molar mass of phosphorus is 123.46 g mol^–1.

Let the volume of  air in the  flask at 27°C be V₁ & that of the same amount of the gas at 477°C be V₂.

According to charle’s law

V₂/T₂  =  V₁/ T₁   …………………(1)

NOW  volume of gas expelled out = V₂ – V₁, THEN

Fraction of the gas expelled out = (V₂ – V₁ ) / V₂ = 1- (V₁/ V₂)     …………….(2)

Also from equation (1)  V₁/ V₂ = T₁/ T₂   …………..(3)

Substituting the values of (3) in (2), we get

Fraction of the air expelled  = 1- T₁/ T₂ = (T₂ – T₁)/ T₂

 = 750- 300 /750 = 0.6

Hence, fraction of air expelled out is 0.6 or 3/5 th

Given,

Amount of the gas, n = 4.0 mol

Volume of the gas, V = 5 dm³

Pressure of the gas, p = 3.32 bar

R = 0.083 bar dm³ K^–1 mol^–1

From the ideal gas equation, we get

pV = nRT

⇒ T = pV / nR

      = (3.32x5) /  (4x0.083)

      = 50K

Hence, the required temperature is 50 K.

Mass of the nitrogen = 1.4 g

Molar mass of nitrogen  = 28 g mol^–1

Therefore amount of nitrogen = 1.4 / 28 = 0.05 mol

Number of nitrogen molecule in 0.05 mol

= 6.023 x 10²³  x 0.05

= 3.0115 x 10²²

Also number of electrons in one molecule of nitrogen  = 14

Therefore, total number of electrons in 1.4 g of nitrogen 

= 3.0115 x 10²² x 14 

= 4.22 x 10²³

We know , Avogadro number = 6.023 × 10²³

Therefore , time taken for distributing the grains

= (6.023 × 10²³ x 1) / 10¹⁰ = 6.023 x 10¹³ s

By Converting these seconds into years, we get

= (6.023 x 10¹³) / (60 x60x24x365)

= 1,90,9800 y

Hence, the time taken would be 1.909 x 10⁶ years.

Given,

Mass of oxygen  = 8 g, molar mass of oxygen = 32 g/mol

Mass of hydrogen = 4 g, molar mass of hydrogen = 2 g/mol

Therefore amount of oxygen = 8/ 32 = 0.25 mol

And amount of hydrogen = 4/2 = 2 mol

From the gas equation PV = n RT, we get,

 P X 1  = (0.25 + 2)  X 0.083 X 300 = 56.02 bar

Hence, the total pressure of the mixture is 56.02 bar.

Payload of the ballon  = mass of the displaced air – mass of the ballon

Radius of the ballon, r  = 10 m

Mass of the ballon, m = 100kg

Therefore volume of the ballon = 4/3πr³ = 4/3 x 22/7 x (10)³ = 4190.5 m³

Now volume of the displaced air = 4190.5 m³

Given,

Density of air = 1.2 kg m^–3

Therefore, the mass of the displaced air

= 4190.5 x 1.2  = 5028.6 kg

Let w be the mass of helium gas filled into the ballon,then

PV = (w/m)RT

OR w = PVM/RT

= (1.66 X 4190.5 X 10³ X 4) / (0.083 X 300)

 = 1117 kg (approx)

Total mass of the balloon filled with He  = 1117 + 100 = 1217 kg

Therefore payload of the balloon  = 5028.6 – 1217 = 3811.6 kg

Hence, the pay load of the balloon is 3811.6 kg.

From the gas equation,

PV = (w/M) RT

Where w is the mass of the gas, M is the molar mass of the gas

For CO₂ , M = 44 g/mol

Substituting the given values ,we get

1 x V  = (8.8 / 44) x 0.083 x 304.1

 = 5.05 L

Hence, the volume occupied is 5.05 L.

From the gas equation,

PV = (w/M) RT

Substituting the given data in the gas equation, we get

PV = (2.9 / M) x R x 368

&

PV = (0.184 / 2) x R x 290

From these two equation, we can write

(2.9 / M) x R x 368 = (0.184 / 2) x R x 290

By, striking throug R from both side, we get

(2.9 / M) x 368 = (0.184 / 2) x 290

Or

(2.9 / M)   =  (0.092 X 290) / 368

Or

M  = 2.9 x 368 / 0.092 x 290

= 40 g/mol

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

Pressure of the gas mixture = 1 bar

Let us consider 100g of the mixture

So ,mass of hydrogen in the mixture  = 20 g

& mass of oxygen in the mixture = 80 g

Using the respective molar masses, we get

  n_H = 20/ 2  = 10 mol

. n_O = 80 / 32 = 2.5 mol

Then, p_H = X_H x P_total

= (n_H / n_{H +} n_O) x P _total  

= (10 / 10 + 2.5 ) x  1 

= 0.8 bar

Hence, the partial pressure of dihydrogen is 0.8 bar.

The SI unit of the given quantity is obtained by substituting the SI units of all the quantities in its expression.

Also

The SI unit for pressure, p is Nm^–2.

The SI unit for volume, V is m^3.

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 given by substituting the above values in the given expression

pV²T² / n  = (Nm^-2) x (m³)² x K² / mol

= Nm⁴K² mol^-1

Charles observed that the volume of certain amount of a gas changes by  V_O / 273.15 for each degree rise or fall in temperature.V_O being the volume at 0°C.The volume at any temperature t°C is given by

V_t  = V_O(1 + t°C/ 273.15)

Now ,if the temperature is lowered, it becomes clear that at t = -273.15°C

V_t  = V_O(1+ (-273.15/273.15))

= V_O  = (1-1) = 0

And below t = -273.15°C, Vt  becomes negative, which is not possible

Therefore the physically significant lowest temperature is -273.15° C

The maximum temperature at which a gas can be converted into a liquid by an increase in pressure is called its critical temperature(T_c). This means that the intermolecular forces of attraction between the molecules of a gas are directly proportional to its critical temperature. Hence, intermolecular forces of attraction are stronger in the case of CO₂.

The van der Waals equation is an equation of state for a fluid composed of particles that have a non-zero volume and a pairwise attractive inter particle force (such as the van der Waals force). The equation is van der Waals equation 

Physical significance of ‘a’: ‘a’ is a measure of the magnitude of intermolecular attractive forces between the particles

Physical significance of ‘b’:

‘b’ is the volume excluded by a mole of particles

  p is the pressure of fluid

V  is the total volume of container containing the fluid