# Question-1

**What will be the minimum pressure required to compress 500 dm**

^{3}of air at 1 bar to 200 dm^{3}at 30^{0 }c?**Solution:**

P

_{1}= 1 bar

P

_{2}= ?

V

_{1}= 500 dm

^{3}

V

_{2}= 200dm

^{3}

P

_{1}V

_{1}= P

_{2}v

_{2}

1 × 500 = P

_{2}× 200 or

P

_{2 }= = 2.5 bar.

# Question-2

**A vessel of 120 mL capacity contains a certain amount of gas at 35 °C and 1.2 bar**

**pressure. The gas is transferred to another vessel of volume 180 mL at 35 °C. What**

**would be its pressure?**

**Solution:**

Let V

_{1 }= 120 ml

V

_{2 }= 180 ml

P

_{1 }= 1.2 bar

P

_{2 }= ?

P

_{1}V

_{1 }= P

_{2}V

_{2}

1.2 × 120 = P

_{2}× 180

P

_{2}=

P

_{2 }= 0.8 bar.

# Question-3

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

**P.**

**Solution:**

PV = nRT

(or) P (Pressure of gas) =

We know = ρ (density)

P = n. ρ RT

Pressure of gas (P) α density of the gas (d).

# Question-4

**At 0**

^{0}C, the density of a gaseous oxide at 2 bar is same as that of nitrogen at 5 bar. What is the molecule mass of the oxide?**Solution:**

Density of nitrogen, ρ =

= P = 5 × 0.987 atm, m = 28

= ρ =

Density of gaseous oxide

=

x =

or x = = 70 g/mol.

# Question-5

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

^{0}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**Solution:**

Amount of ideal gas, A = W

_{A }= 1 g

Its pressure, P

_{A}= 2bar

Amount of another gas, B = W

_{B }=

_{ }2 g

Total pressure of the flask = 3 bar

Partial Pressure of gas B in the gas mixture = 3-2 = 1 bar

PV = nRT

=

For gas A, P

_{A}=

2 = ………(i)

For gas B,

P

_{B }=

1 = --------(ii)

Divide equation (i) by (ii)

or = 2

4M

_{A}= M

_{B}

The molecular mass of M

_{b}is four times of M

_{a}.

# Question-6

**The drain cleaner, Drainex contains small bits of aluminium which react with caustic soda to produce hydrogen. What volume of hydrogen at 20**

^{0}c and one bar will be released when 0.15 g of aluminium reacts?**Solution:**

2Al + 2NaOH + 2H

_{2}O → 2NaAlO

_{2}+ 3H

_{2}

2 × 27 g of Al 3 × 22.4 litre of H

_{2}

= 54 g

54g Al gives = 3 × 22.4 litre H

_{2}at N.T.P.

0.15 g Al gives = = 0.186 litre at N.T.P.

=

Normal Pressure P

_{1}= 1 atm

P

_{2}= 1 bar = 0.987 atm

V

_{1 }= 0.186L

V

_{2}=?

T_{1} = 273K

T_{2} = 273 + 20 = 293K

=

V_{2} = = 0.202 litre = 202 ml.

# Question-7

**What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm**

^{3}flask at 27^{0}C?**Solution:**

= 3.2 g

Mol. Weight of CH

_{4}= 16

Moles of CH

_{4 }= =0.2

= 4.4 g

Mol. Weight of CO

_{2}= 44

Moles of CO

_{2}= = 0.1 mol

Total moles present in mixture = 0.2 + 0.1 = 0.3 mol

Volume of flask = 9 dm

^{3}= 9 L

Temperature = 273 + 27 = 300 K

Now for pressure of mixture = P × V = nRT

P × 9 = 0.3 × 0.0821 × 300

P = =0.821 atom

0.821 × 1.01 × 10

^{5}P

_{a}= 8.29 × 10

^{4}P

_{a}.

# Question-8

**What will be the pressure of the gas mixture when 0.5 L of H**

_{2}at 0.8 bars and 2.0 L of oxygen at 0.7 bar are introduced in a 1 L vessel are 27^{0}C?**Solution:**

Volume of H

_{2}, (V

_{1}) 0.5 l

Pressure of H

_{2}, (P

_{1}) = 0.8 bar

Volume of oxygen, (V

_{2}) = 2.0 l

Pressure of O

_{2}, (P

_{2}) = 0.7 bar

Total Volume of the flask (V

_{1 }+ V

_{2}) = 1l

Total pressure of the gaseous mixture is P

P (V

_{1 }+ V

_{2}) = P

_{1}V

_{1}+ P

_{2}V

_{2}

P × 1 = (0.8 × 0.5) + (0.7 × 2.0)

P × 1 = (0.4 + 1.4)

Total Pressure of the mixture of gases , P = 1.8/1 = 1.8 bar.

# Question-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?**Solution:**

d

_{1}= 5.46 g/dm

^{3}d

_{2}= ?

T

_{1}= 273 + 27 = 300 K T

_{2 }= 273 K

P

_{1}= 2 bar

=

d

_{2}= = 3g/dm

^{3}.

# Question-10

**34.05 mL of phosphorous vapour weighs 0.00625 g at 546**

^{o }C and 0.1 bar pressure. What is the molar mass of phosphorous?**Solution:**

P = 0.1 bar , V = 34.05 mL = 0.03405

R = 0.0821 litre-atm K

^{-1}mol

^{-1}

T = 273 + 546 = 819 K

PV = nRT

n = =

= 0.00005 mol

^{-1 }

Weight of phosphorous vapour = 0.00625g

Molar mass of phosphorus = = 125 g.

# Question-11

**A student forgot to add the reaction mixture to the round bottomed flask at 27**

^{o}C but put it on the flame. After a lapse of time, he realized his mistake, using a pyrometer he found the temperature of the flask was 477^{o}C. What fraction of air would have been expelled out?**Solution:**

or

T

_{1}= 27+273=300 K

T

_{2 }= 477 + 273 = 750 K

=

If the total volume of flask is taken as 1

Fraction of air expelled = 1 - =.

# Question-12

**Calculate the temperature of 4.0 moles of a gas occupying 5 dm**

^{3}at 3.32 bar (R=0.083 bar dm^{3}K^{-1}mol^{-1}).**Solution:**

PV = nRT

P = 3.32 bar, V = 5dm

^{3}

n = 4 moles

R = 0.083 bar dm

^{3}K

^{-1}mol

^{-1}

∴ 3.32 × 5 = 4 × 0.083 × T

T = = 50K.

# Question-13

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

**Solution:**

Atomic weight of Nitrogen = 14 g

weight of Nitrogen gas = 1.4 g

Number of gram Moles of Nitrogen atoms = = = 0.1 mol.

No. of atoms of Nitrogen, present in 1 mol of Nitrogen gas = 6.022 × 10

^{23}atoms

1 atom of Nitrogen contains = 7 electrons

No. of electrons, present in 1 mol of Nitrogen atoms = 7 × 6.022 × 10

^{23}

No. of electrons in 0.1 mol = 0.1 × 7 × 6.02 × 10

^{23}= 4.22 × 10

^{23}.

# Question-14

**How much time it would take to distribute one Avogadro number of wheat grains if 10**

^{10}grains are distributed each second?**Solution:**

10

^{10}grains are distributed I s Time in seconds, taken to distribute

6.023 × 10^{23} grains = s = years = 1,90,800 years.

# Question-15

**Calculate the total pressure in a mixture of 8 g of oxygen and 4 g of hydrogen confined in a vessel of 1 dm**

^{3}at 27^{o}C, R = 0.083 bar dm^{3}K^{-1}mol^{-1}.**Solution:**

Weight of oxygen = 8 g

Moles of O

_{2}= =

= 4 g

Moles of H

_{2}= =2

Total moles present in mixture = =

Volume of vessel = 1 dm

^{3}= 1 L

Temperature = 27

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

Now for pressure of mixture = P × V = n × RT

P × 1 =

P = 56.025 bar.

# Question-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 10m, mass 100kg 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}).**Solution:**

Volume of balloon, V =

r = 10 m

V = =

Volume of air displaced = Volume of balloon = 4187 m

^{3}

M = Vol × Density

Mass of displaced air = 4187 m

^{3}× 1.2 kg m

^{-3}= 5024. 4 kg

Temp, T = 27+273=300 K

Moles of gas present, n =

= =mol.

Mass = No. of moles × Molecular weight

Mass of He present = 279 × 10^{3} × 4

= 1116.4 × 10^{3} g

= 1116.4 kg

Mass of filled balloon = 100 + 1116.4 = 1216.4 Kg

Pay load = Mass of displaced air - Mass of balloon

= 5024.4 - 1216.4

= 3808 kg.

# Question-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 1 K^{-1}mol^{-1}.**Solution:**

P = 1 bar , T = 31.1 + 273 = 304.1K

Mol. wt of CO_{2} = 44

Weight of CO_{2} = 8.8 g

PV = (W/m)RT

I x V =

V = 5.05 L.

# Question-18

**2.9 g of a gas at 95**

^{o}C occupied the same volume as O.184 g of hydrogen at 17^{o}C at the same pressure. What is the molar mass of the gas?**Solution:**

Ideal gas equation, PV = nRT

**1**

^{st}CaseLet the molar mass of gas (M) = M(g mol

^{-1})

Mass of gas (m) = 2.9 g

No. of moles (m) = =mol

T = 273 + 95 = 368

Pressure = P, Volume = V

PV = nRT

PV = ………(i)

**2**

^{nd}CaseMass of hydrogen = 0.184 g

Number of moles of H

_{2}= mol

T = 273 + 17 = 290

PV = ………..(ii)

In both gases PV is same

From (i) and (ii)

=

M = = 40 g mol

^{-1}.

# Question-19

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

**Solution:**

Percentage of H

_{2}= 20%, say 20g

No. of moles of H

_{2}= = 10 moles

Mass of oxygen = (100-20) = 80 g)

No. of moles of O

_{2}= = 2.5 moles

Mole fraction of H

_{2}= =

Partial pressure of H

_{2}=Total Pressure × Mole fraction of H

_{2}

Total Pressure = 1 bar

= 1 × = 0.8 bar.

# Question-20

**What would be the SI unit for the quantity pV**

^{2}T 2/n ?**Solution:**

S.I. unit = = Pa m

^{6}K

^{2}mol

^{-1}.

# Question-21

**In terms of Charles’ law explain why –273°C is the lowest possible temperature.**

**Solution:**

Charles law states that Pressure, remaining constant the volume of a given mass of gas increases or decreases by 1/273 of its volume at 0

^{0}c for every 1

^{0}c rise or fall in temperature .

V

_{t}= V

_{0}(1 + )

If t = -273

^{0}c

V

_{t}= V

_{0 }(1 - )

= 0

Thus at -273

^{0}c any volume of a given mass of gas becomes zero. Thus -273

^{0}c is said to be limiting temperature and it is equal to OK (absolute scale).

# Question-22

**Critical temperature for carbon dioxide and methane are 31.1°C and –81.9°C respectively. Which of these has stronger intermolecular forces and why?**

**Solution:**

CO

_{2}can more easily be liquified than Methane which shows that intermolecular forces in CO

_{2}are more stronger than in Methane, because in CO

_{2}polarity exists.

# Question-23

**Explain the physical significance of van der Waals parameters.**

**Solution:**

Vander Waals equation of state for mol of gas = (v-b) = RT

For n moles of gas = (v-nb) = RT

The constant 'a' tells the magnitude of attractive forces between the molecules of the gas. In other words larger the value of 'a', greater the intermolecular attraction.

The constant 'b' relates to non compressible volume of gas, as gas molecules occupy some volume, however small may be, at high pressures and low temperatures.∴Actual volume of gas = Available volume - Non compressible volume = (V - b).