Class 11 - Chemistry - States of Matter

Question 5.1

What will the minimum pressure required to compress 500 dm^{3} of air at 1 bar to 200 dm^{3} at 30^{o} C?

Answer.

P_{1} = 1 bar

V_{1} = 500 dm^{3}, V_{2} = 200 dm^{3}

As temperature remains constant at 30^{o} C,

P_{1}V_{1} = P_{2}V_{2}

1 bar x 500 dm^{3} = P_{2} x 200 dm^{3}

P_{2} = (500/ 200) = 2.5 bar

Question 5.2

A vessel of 120 mL capacity contains a certain amount of gas at 35^{o} C and 1.2 bar pressure.

The gas is transferred to another vessel of volume 180 mL at 35^{o} C. What would be its pressure?

Answer.

V_{1} = 120 mL, V_{2} = 180 mL

P_{1} = 1.2 bar

As temperature remains constant, P_{1}V_{1} = P_{2}V_{2}

1.2 bar x 120 mL = P_{2} x 180 mL

P_{2} = 0.8 bar

Question 5.3

Using the equation of state pV = nRT, show that at a given temperature density of a gas is proportional to gas pressure p.

Answer.

According to ideal gas equation,

PV = nRT

PV = (nRT/V)

n = (Constant Mass of gas/Molar mass of gas)

P = (mRT/MV) [ρ(density) = (m/V)]

P = (ρRT/M)

P ∝ ρ [at constant temperature]

Question 5.4

At 0^{o}C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar.

What is the molecular mass of the oxide?

Answer.

Using expression, d = MP/RT at same temperature and for same density.

M_{1}P_{1} = M_{2}P_{2} [because R is constant]

M_{1} x 2 = 28 x 5 [molecular mass of nitrogen = 28 u]

M_{1} = 70 u

Question 5.5

Pressure of 1 g of an ideal gas A at 27^{o}C is found to be 2 bar.

When 2 g of another ideal gas B is introduced in the same flask at same temperature the pressure becomes 3 bar.

Find a relationship between their molecular masses.

Answer.

For ideal gas A, ideal gas equation is

P_{A}V = n_{A}RT -----(1)

Where p_{A} represent the pressure of gas A.

and n_{A} represents and number of moles of gas A.

For ideal gas B, ideal gas equation is given by

P_{B}V = n_{B} RT ----(2)

P_{B} represents pressure of gas B.

N_{B} represents pressure of gas B.

P_{A}V = (m_{A}/M_{A}) RT

(P_{A}M_{A}/m_{A})= (RT/V) -----(3)

From equation (2)

P_{B}V = (m_{B}/M_{B}) RT

(P_{B}M_{B}/m_{B} )= (RT/V) -----(4)

M_{A }and M_{B }are the molecular masses of gases A and B.

(P_{A}M_{A}/m_{A})= (P_{B}M_{B}/m_{B}) ------(5)

m_{A }= 1g

P_{A} = 2 bar

M_{B} = 2 g

P_{B} = (3 – 2) = 1 bar [total pressure is 3 bar]

Substituting valves in equation (5),

(2 x M_{A})/1 = (1 x M_{B})/ 2

4 M_{A }= M_{B}

Question 5.6

The drain cleaner, Drainex contains small bits of aluminium which react with caustic soda to produce dihydrogen.

What volume of dihydrogen at 20^{o} C and one bar will be released when 0.15 g of aluminum reacts?

Answer.

2Al + 2NaOH + 2H_{2}O --> 2NaAlO_{2} + 3H_{2}

At standard temperature and pressure (273.15 K and 1 atm)

54g ( 2 x 27 g) of Al gives 3 x 22400 mL of H_{2}.

0.15 g Al = (3 x 22400 x 0.15)/54 mL of H_{2}

= 186.67 mL of H_{2}

At standard temperature and pressure,

P_{1} = 1 atm

V_{1} = 186.67 mL

T_{1} = 273.15 K

Let us consider the volume of dihydrogen = V_{2} when P_{2} = 0.987 atm

T2 = 20^{o} C = (273.15 + 20) K = 293.15 K

(P_{1}V_{1}/T_{1}) = (P_{2} V_{2}/T_{2})_{ }

V_{2} = (P_{1}V_{1}T_{2}/P_{2}T_{1} )= (1 x 186.67 x 293.15)/(0.987 x 273.15) = 202.98 mL

= 203 mL

Question 5.7

What will be the pressure exerted by a mixture 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm^{3} flask at 27^{o} C?

Answer.

P = nRT/V = mRT/MV

For methane, CH_{4}

pCH_{4} = (3.2 x 8.314 x 300)/(16 x 9 x 10^{-3}) = 5.543 x 10^{4} Pa

Here, 9 dm^{3 }= 9 x 10^{-3} m^{3} and 27^{o} C = 300 K)

For Carbon dioxide, CO_{2}

pCO_{2}

= (4.4/44) x (8.314 x 300)/(9 x 10^{-3}) = 2.771 x 10^{4} Pa

Total pressure exerted by the mixture, P = pCH_{4} + pCO_{2}

P = (5.543 x 10^{4} + 2.771 x 10^{4}) = 8.314 x 10^{4} Pa

Question 5.8

What will be the pressure of the gaseous mixture when 0.5 L of H_{2} at 0.8 bar and 2.0 L of dioxide at 0.7 bar are introduced in a 1L vessel at 27^{o}C?

Answer.

Partial pressure of H_{2} in 1L vessel

P_{1} = 0.8 bar, V_{1} = 0.5 L, V_{2} = 1.0 L

As temperature remains constant, P_{1}V_{1} = P_{2}V_{2}

(0.8 bar) (0.5 L) = P_{2} (1.0 L)

P_{2} = (1.0L)

P_{2} = 0.40 bar

PH_{2} = 0.40 bar

Partial pressure of oxygen in 1 L vessel.

P_{1}’V_{1}’ = P_{2}’V_{2}’

(0.7 bar) (2.0L) = P_{2}’ (1 L)

P_{2}’ = 1.4 bar

Total pressure = P_{H2} + P_{O2} = 0.4 bar + 1.4 bar = 1.8 bar

Question 5.9

Density of a gas is found to be 5.46 g/dm^{3} at 27^{o} C at 2 bar pressure. What will be its density at STP?

Answer.

d_{1} = 5.46 g/dm^{3}

p_{1} = 2 bar

T_{1} = 27^{o} C = 27 + 273 = 300 K

p_{2 }= 1 bar

T_{2} = 273 K

Density d_{2} of the gas at STP

5.46 g dm^{-3}/d_{2} = (2 bar/273K)/(300 K x 1 bar)

d_{2} = 3 g dm^{-3}

Question 5.10

34.05 mL of phosphorous vapour weighs 0.625 g at 546^{o} C and 0.1 bar pressure.

What is the molar mass of phosphorus?

Answer.

p = 0.1 bar

V = 34.05 mL = 34 x 10^{-3} dm^{3}

R = 0.083 bar dm^{3} at K^{-1} mol^{-1}

T = 546^{o} C = 546 + 273 = 819 K

pV = nRT

n = pV/RT = (0.1 x 34.05 x 10^{-3})/(0.083 x 819) = 5.01 x 10^{-5} mol

Molar mass of Phosphorus = 0.0625/ (5.01 x 10^{-5}) = 1247.5 g mol^{-1}.

Question 5.11

A student forgot to add the reaction mixture to the round bottomed flask at 27^{o} C but instead he/she placed the flask on the flame.

After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was 477 ^{o} C.

What function of air would have been expelled out?

Answer.

Volume of air in the flask at 27^{o} C = V cm^{3}

(V_{1}/T_{1}) =(V_{2}/T_{2})

(V/300) = (V_{2}/750)

V_{2} = 2.5 V

Volume expelled = 2.5 V – V = 1.5 V

Fraction of air expelled = (1.5 V /2.5V) =(3/5)

Question 5.12

Calculate the temperature of 4.0 mol of a gas occupying 5 dm^{3} at 3.32 bar. (r = 0.083 bar dm^{3} K^{-1} mol^{-1}).

Answer.

PV = nRT

T = PV/nR

T = __(__3.32 bar x 5 dm^{3 )}/(4.0 mol x 0.083 bar dm^{3} )K^{-1} mol^{-1}

T = 50 K

Question 5.13

Calculate the total number of electrons present in 1.4 g of dinitrogen gas.

Answer.

Molecular mass of nitrogen = 28 g

1.4 g of nitrogen gas = (1.4/28) = 0.05 mol

= 0.05 x 6.02 x 10^{23} number of molecules

= 3.01 x 10^{23} number of molecules

One molecule of nitrogen has 14 electrons.

So, 3.1 x 10^{23} molecules of Nitrogen contains = 14 x 3.01 x 10^{23}

= 4.214 x 10^{23} electrons

Question 5.14

How much time would it take to distribute one Avogadro number of wheat grains, if 10^{10} grains are distributed each second?

Answer.

Time taken = (6.02 x 10^{23})/10^{10} s = 6.02 x 10^{13} s

Time taken = 6.02 x 10^{23}/(60 x 60 x 24 x 365) years = 1.909 x 10^{6} years.

Question 5.15

Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a

vessel of 1 dm^{3} at 27^{o} C. R = 0.083 bas dm^{3} K^{-1} mol^{-1}.

Answer.

Molar mass of oxygen = 32 g mol^{-1}

8 g O_{2} = 8/32 mol = 0.25 mol

Molar mass of hydrogen = 2 g mol^{-1}

4 g of H_{2} = 4/2 = 2 mol

Total number of moles (n) = 2 + 0.25 = 2.25

V = 1 dm^{3}

T = 27^{o} C = 300 K

R = 0.083 bar dm^{3} K^{-1} mol^{-1}

PV = nRT

P = nRT/V

Putting all the values, we get

P = 56.025 bar

Question 5.16

Pay load is defined as the difference between the mass of displaced air and the mass of the balloon.

Calculate the pay load when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at 27^{o} C.

(Density of air = 1.2 kg m^{-3} and R = 0.083 bar dm^{3} K^{-1} mol^{-1}.

Answer.

Volume of balloon = 4/3 π r^{3} = (4/3) x (22/7) x (10)^{3} = 4190.5 m^{3}

Radius of balloon = 10 m

Volume of Helium filled at 1.66 bar and 27^{o}C = 4190.5 m^{3}

Mass of Helium, PV = nRT = wRT/M

W = MPV/RT = (4 x 10^{-3} kg mol^{-1}) (1.66 bar) (4.190.5 x 10^{3} dm^{3}) / ((0.083 bar dm^{3} K^{-1} mol^{-1}) (300 K))= 1117.5 kg

Total mass of balloon with Helium = (100 + 1117.5) = 1217.5 kg

Maximum mass of air that can be displaced by balloon to go up = (Volume x density)

= 4190.5 m^{3} x 1.2 kg m^{-3} = 5028.6 kg

Pay load = 5028.6 – 1217.5 kg = 3811.1 kg

Question 5.17

Calculate the volume occupied by 8.8 g of CO_{2} at 31.1^{o} C and 1 bar pressure. R = 0.083 bar L K^{-1} mol^{-1}.

Answer.

Number of moles of carbon dioxide (n) = Mass of carbon dioxide/Molar mass

n = 8.8g/44 g mol^{-1}

Pressure of carbon dioxide (P) = 1 bar

R = 0.083 bar L K^{-1} mol^{-1}

Temperature, t = 273 + 31.1 = 304.1 K

PV = nRT

V = (nRT/P) = (0.2 x 0.083 x 304.1)/1

V = 5.048 L

Question 5.18

2.9 g of a gas at 95^{o} C occupied the same volume as 0.184 g of dihydrogen at 17^{o} C, at the same pressure.

What is the molar mass of the gas?

Answer.

V = (mRT/Mp) = (0.184 x R x 290)/ (2 x p)

Let M is the molar mass of unknown gas

Volume occupied by unknown gas = (mRT/Mp)

V = (2.9 x R x 368)/ (M x p)

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

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

Molar mass of gas = 40 g mol^{-1}

Question 5.19

A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen.

Calculate the partial pressure of dihydrogen.

Answer.

n_{H2} = (20/2) = 10 moles

n_{O2} = (80/32) = 2.5 moles

p_{H2} = (n_{H2} )x P_{total }

n_{H2} + n_{O2}

Putting all the values, p_{H2} = 0.8 bar

Question 5.20

What would be the SI unit for the quantity (pV^{2}T^{2}/n)?

Answer.

__(__Nm^{-2}) (m^{3})^{2} (K^{2}) / (mole)= Nm^{4} K^{2} mol^{-1}

Question 5.21

In terms of Charles’ law explain why -273^{o} C is the lowest possible temperature.

Answer.

At -273^{o} C, volume of gas becomes equal to zero.

This happens because all gasses get transferred into liquid form before reaching -273^{∘} C.

Therefore, it can be said that −273^{∘}C is the lowest possible temperature.

Question 5.22

Critical temperature for carbon dioxide and methane are 31.1^{o} C and -81.9^{o} C respectively.

Which of these has stronger intermolecular forces and why?

Answer.

Higher the critical temperature, more easily the gas can be liquefied.

That is the intermolecular forces of attraction among the molecules of gas are directly proportional to its critical temperature.

So, CO_{2} has stronger intermolecular forces than CH_{4.}

Question 5.23

Explain the physical significance of van der Waals parameters?

Answer.

Physical significance ‘a’ represents the magnitude of intermolecular attractive forces within gas.

Physical significance ‘b’ represents volume of gas molecule.

.