Class 11 - Chemistry - States of Matter

Question 5.1

What will the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30o C?

Answer.

P1 = 1 bar

V1 = 500 dm3, V2 = 200 dm3

As temperature remains constant at 30o C,

P1V1 = P2V2

1 bar x 500 dm3 = P2 x 200 dm3

P2 = (500/ 200) = 2.5 bar

Question 5.2

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

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

Answer.

V1 = 120 mL, V2 = 180 mL

P1 = 1.2 bar

As temperature remains constant, P1V1 = P2V2

1.2 bar x 120 mL = P2 x 180 mL

P2  = 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 0oC, 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.

M1P1 = M2P2 [because R is constant]

M1 x 2 = 28 x 5 [molecular mass of nitrogen = 28 u]

M1 = 70 u

Question 5.5

Pressure of 1 g of an ideal gas A at 27oC 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

PAV = nART -----(1)

Where pA represent the pressure of gas A.

and nA represents and number of moles of gas A.

For ideal gas B, ideal gas equation is given by

PBV = nB RT ----(2)

PB represents pressure of gas B.

NB represents pressure of gas B.

PAV = (mA/MA) RT

(PAMA/mA)= (RT/V) -----(3)

From equation (2)

PBV = (mB/MB) RT

(PBMB/mB )= (RT/V) -----(4)

MA and MB are the molecular masses of gases A and B.

(PAMA/mA)= (PBMB/mB)   ------(5)

mA = 1g

PA = 2 bar

MB = 2 g

PB = (3 – 2) = 1 bar [total pressure is 3 bar]

Substituting valves in equation (5),

(2 x MA)/1 = (1 x MB)/ 2

4 MA = MB

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 20o C and one bar will be released when 0.15 g of aluminum reacts?

Answer.

2Al + 2NaOH + 2H2O --> 2NaAlO2 + 3H2

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

54g ( 2 x 27 g) of Al gives 3 x 22400 mL of H2.

0.15 g Al = (3 x 22400 x 0.15)/54 mL of H2

= 186.67 mL of H2

At standard temperature and pressure,

P1 = 1 atm

V1 = 186.67 mL

T1 = 273.15 K

Let us consider the volume of dihydrogen = V2 when P2 = 0.987 atm

T2 = 20o C = (273.15 + 20) K = 293.15 K

(P1V1/T1) = (P2 V2/T2)

V2 = (P1V1T2/P2T1 )= (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 dm3 flask at 27o C?

Answer.

P = nRT/V = mRT/MV

For methane, CH4

pCH4 = (3.2 x 8.314 x 300)/(16 x 9 x 10-3) = 5.543 x 104 Pa

Here, 9 dm3 = 9 x 10-3 m3 and 27o C = 300 K)

For Carbon dioxide, CO2

pCO2

= (4.4/44) x (8.314 x 300)/(9 x 10-3) = 2.771 x 104 Pa

Total pressure exerted by the mixture, P = pCH4 + pCO2

P = (5.543 x 104 + 2.771 x 104) = 8.314 x 104 Pa

Question 5.8

What will be the pressure of the gaseous mixture when 0.5 L of H2 at 0.8 bar and 2.0 L of dioxide at 0.7 bar are introduced in a 1L vessel at 27oC?

Answer.

Partial pressure of H2 in 1L vessel

P1 = 0.8 bar, V1 = 0.5 L, V2 = 1.0 L

As temperature remains constant, P1V1 = P2V2

(0.8 bar) (0.5 L) = P2 (1.0 L)

P2 = (1.0L)

P2 = 0.40 bar

PH2 = 0.40 bar

Partial pressure of oxygen in 1 L vessel.

P1’V1’ = P2’V2

(0.7 bar) (2.0L) = P2’ (1 L)

P2’ = 1.4 bar

Total pressure = PH2 + PO2 = 0.4 bar + 1.4 bar = 1.8 bar

Question 5.9

Density of a gas is found to be 5.46 g/dm3 at 27o C at 2 bar pressure. What will be its density at STP?

Answer.

d1 = 5.46 g/dm3

p1 = 2 bar

T1 = 27o C = 27 + 273 = 300 K

p2 = 1 bar

T2 = 273 K

Density d2 of the gas at STP

5.46 g dm-3/d2 = (2 bar/273K)/(300 K x 1 bar)

d2 = 3 g dm-3

Question 5.10

34.05 mL of phosphorous vapour weighs 0.625 g at 546o 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 dm3

R = 0.083 bar dm3 at K-1 mol-1

T = 546o 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 27o 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 27o C = V cm3

(V1/T1) =(V2/T2

(V/300) = (V2/750)

V2 = 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 dm3 at 3.32 bar. (r = 0.083 bar dm3 K-1 mol-1).

Answer.

PV = nRT

T = PV/nR

T = (3.32 bar x 5 dm3 )/(4.0 mol x 0.083 bar dm3 )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 1023 number of molecules

= 3.01 x 1023 number of molecules

One molecule of nitrogen has 14 electrons.

So, 3.1 x 1023 molecules of Nitrogen contains = 14 x 3.01 x 1023

= 4.214 x 1023 electrons

Question 5.14

How much time would it take to distribute one Avogadro number of wheat grains, if 1010 grains are distributed each second?

Answer.

Time taken = (6.02 x 1023)/1010 s = 6.02 x 1013 s

Time taken = 6.02 x 1023/(60 x 60 x 24 x 365) years = 1.909 x 106 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 dm3 at 27o C. R = 0.083 bas dm3 K-1 mol-1.

Answer.

Molar mass of oxygen = 32 g mol-1

8 g O2 = 8/32 mol = 0.25 mol

Molar mass of hydrogen = 2 g mol-1

4 g of H2 = 4/2 = 2 mol

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

V = 1 dm3

T = 27o C = 300 K

R = 0.083 bar dm3 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 27o C.

(Density of air = 1.2 kg m-3 and R = 0.083 bar dm3 K-1 mol-1.

Answer.

Volume of balloon = 4/3 π r3 = (4/3) x (22/7) x (10)3 = 4190.5 m3

Radius of balloon = 10 m

Volume of Helium filled at 1.66 bar and 27oC = 4190.5 m3

Mass of Helium, PV = nRT = wRT/M

W = MPV/RT = (4 x 10-3 kg mol-1) (1.66 bar) (4.190.5 x 103 dm3) / ((0.083 bar dm3 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 m3 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 CO2 at 31.1o 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 95o C occupied the same volume as 0.184 g of dihydrogen at 17o 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.

nH2 = (20/2) = 10 moles

nO2 = (80/32) = 2.5 moles

pH2 = (nH2 )x Ptotal

nH2 + nO2

Putting all the values, pH2 = 0.8 bar

Question 5.20

What would be the SI unit for the quantity (pV2T2/n)?

Answer.

(Nm-2) (m3)2 (K2)  / (mole)= Nm4 K2 mol-1

Question 5.21

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

Answer.

At -273o 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 −273C is the lowest possible temperature.

Question 5.22

Critical temperature for carbon dioxide and methane are 31.1o C and -81.9o 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, CO2 has stronger intermolecular forces than CH4.

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.

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