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Question-1

What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3  at 300 c?

Solution:

P1 = 1 bar

P2 = ?  

V1 = 500 dm3
 

V2 = 200dm3  

P1V1 = P2v2  

1 × 500 = P2 × 200 or

P2 = = 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 V1 = 120 ml

V2 = 180 ml

P1 = 1.2 bar

P2 = ?


P1V1 = P2V2

1.2 × 120 = P2 × 180

P2 =

P2 = 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 00 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 270 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 = WA = 1 g
Its pressure, PA = 2bar
Amount of another gas, B = WB = 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, PA
=
2 = ………(i)
For gas B,
PB =
1 = --------(ii)
Divide equation (i) by (ii)

or = 2
4MA = MB
The molecular mass of Mb is four times of Ma.

Question-6

The drain cleaner, Drainex contains small bits of aluminium which react with caustic soda to produce hydrogen. What volume of hydrogen at 200c and one bar will be released when 0.15 g of aluminium reacts?

Solution:
2Al + 2NaOH + 2H2O 2NaAlO2 + 3H2

2
× 27 g of Al                 3 × 22.4 litre of H2
= 54 g

54g Al gives = 3
× 22.4 litre H2 at N.T.P.

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


=

Normal Pressure P1 = 1 atm             

P2 = 1 bar = 0.987 atm

V1 = 0.186L                                    

V2 =?

T1 = 273K                                      

T2 = 273 + 20 = 293K

=

V2 = = 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 dm3 flask at 270C?

Solution:
= 3.2 g

Mol. Weight of CH4 = 16

Moles of CH4 = =0.2

= 4.4 g

Mol. Weight of CO2 = 44

Moles of CO2 = = 0.1 mol

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

Volume of flask = 9 dm3 = 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 × 105 Pa = 8.29 × 104 Pa.

Question-8

What will be the pressure of the gas mixture when 0.5 L of H2 at 0.8 bars and 2.0 L of oxygen at 0.7 bar are introduced in a 1 L vessel are 270C?

Solution:
Volume of H2, (V1) 0.5 l

Pressure of H2, (P1) = 0.8 bar

Volume of oxygen, (V2) = 2.0 l

Pressure of O2, (P2) = 0.7 bar

Total Volume of the flask (V1 + V2) = 1l

Total pressure of the gaseous mixture is P

P (V1 + V2) = P1V1 + P2V2

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/dm3 at 27oC at 2 bar pressure. What will be its density at STP?

Solution:


d1 = 5.46 g/dm3                 d2 = ?

T1 = 273 + 27 = 300 K        T2
273 K

P1 = 2 bar

=

d2 = = 3g/dm3.

Question-10

34.05 mL of phosphorous vapour weighs 0.00625 g at 546o 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 27oC 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 477oC. What fraction of air would have been expelled out?

Solution:


or

T1 = 27+273=300 K

T2 = 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 dm3 at 3.32 bar (R=0.083 bar dm3 K-1 mol-1).

Solution:
PV = nRT

P = 3.32 bar, V = 5dm3

n = 4 moles

R = 0.083 bar dm3 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 × 1023 atoms

1 atom of Nitrogen contains = 7 electrons

No. of electrons, present in 1 mol of Nitrogen atoms = 7 × 6.022 × 1023

No. of electrons in 0.1 mol = 0.1 × 7 × 6.02 × 1023 = 4.22 × 1023.

Question-14

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

Solution:
1010 grains are distributed I s Time in seconds, taken to distribute

6.023 × 1023 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 dm3 at 27oC, R = 0.083 bar dm3 K-1 mol-1.

Solution:
Weight of oxygen = 8 g

Moles of O2 = =

= 4 g

Moles of H2 = =2

Total moles present in mixture = =

Volume of vessel = 1 dm3 = 1 L

Temperature = 27oC = 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 27oC. (Density of air 1.2 kg m-3 and R = 0.083 bar dm3 K-1 mol-1).

Solution:
Volume of balloon, V =
                              r = 10 m
                              V = =

Volume of air displaced = Volume of balloon = 4187 m3

M = Vol × Density

Mass of displaced air = 4187 m3 × 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 × 103 × 4
                               = 1116.4 × 103 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 CO2 at 31.1oC 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 CO2 = 44

Weight of CO2 = 8.8 g

PV = (W/m)RT

I x V =

V = 5.05 L.

Question-18

2.9 g of a gas at 95oC occupied the same volume as O.184 g of hydrogen at 17oC at the same pressure. What is the molar mass of the gas?

Solution:
Ideal gas equation, PV = nRT

1st Case
Let 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)

2nd Case
Mass of hydrogen = 0.184 g

Number of moles of H2 = 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 H2 = 20%, say 20g

No. of moles of H2 = = 10 moles

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

No. of moles of O2 = = 2.5 moles

Mole fraction of H2 = =

Partial pressure of H2 =Total Pressure × Mole fraction of H2

Total Pressure = 1 bar
                     = 1 × = 0.8 bar.

Question-20

What would be the SI unit for the quantity pV2T 2/n ?

Solution:
S.I. unit = = Pa m6K2mol-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 00c for every 10c rise or fall in temperature .

Vt = V0 (1 + )

If t = -2730c

Vt = V0 (1 - )
    = 0

Thus at -2730c any volume of a given mass of gas becomes zero. Thus -2730c 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:
CO2 can more easily be liquified than Methane which shows that intermolecular forces in CO2 are more stronger than in Methane, because in CO2 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).




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