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```Solved Problems

Question 1:

A gas occupies one litre under atmospheric pressure. What will be the volume of the same amount of gas under 750 mm of Hg at the same temperature?

Solution.

Given V1 = 1 litre

P1 = 1 atm

V2 = ?

P2 = 750 / 760 atm

Using

P1V1 = P2V2

We get

1 × 1 = 750 / 760 × V2

V2 = 1.0133 litre = 1013.3 ml

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Question 2:

How large a balloon could you fill with 4g of He gas at 22°C and 720 mm of Hg?

Solution.

Given, P = 720 / 760 atm, T = 295K, w = 4g

and m = 4 for He

PV = w / M RT

= 720 / 760 × V = 4/4 × 0.0821 × 295

∴ V = 25.565 litre

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Question 3:

Calculate the temperature at which 28g N2 occupies a volume of 10 litre at

2.46 atm

Solution.

w = 28g, P = 2.46 atm, V = 10 litre, m = 28

Now, PV = w / M RT (R = 0.0821 litre atm K–1 mole–1)

T = 299.6 K

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Question 4:

A gas occupies 300 ml at 27°C and 730 mm pressure what would be its volume at STP

Solution

V2 = 300 / 1000 litre, P2 = 730 / 760 atm, T2 = 300K

At STP, V1= ? P1= 1 atm, T1= 273K

P2V2 / T2 = P1V1 / T1 or V1 = 0.2622 litre

Volume at STP = 262.2 ml

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Question 5:

In Victor Meyer’s experiment, 0.23g of a volatile solute displaced air which measures 112 ml at NTP. Calculate the vapour density and molecular weight of the substance.

Solution

Volume occupied by solute at NTP = Volume of air displaced at NTP   = 112 ml

For volatile solute       PV = w / M RT

at NTP, P = 1 atm, T = 273 K

1 × 112/ 1000 = 0.23 / m × 0.0821 × 273

m = 46.02 and V.D. = 23.01

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Question 6:

A cylindrical balloon of 21 cm diameter is to be filled up with H2 at NTP from a cylinder containing the gas at 20 atm at 27°C. The cylinder can hold 2.82 litre of water at NTP. Calculate the number of balloons that can be filled up.

Solution

Volume of 1 balloon which has to be filled = 4/3 π (21/2)3 = 4851 ml   = 4.851 litre

Let n balloons be filled, then volume of H2 occupied by balloons = 4.851 × n

Also, cylinder will not be empty and it will occupy volume of H2 = 2.82 litre.

∴ Total volume occupied by H2 at NTP = 4.851 × n + 2.82 litre

∴ At STP

P2 = 1 atm   Available H2

V1= 4.851 × n + 2.82            P2 = 20 atm

T1 = 273 K   T2 = 300K

P1V1 / T2 = P2V2 / T2  V2 =  2.82 litre

or 1 × (4.85 1n + 2.82 / 273) = 20 × 2.82 / 300 ∴ n = 10

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Question 7:

A 20g chunk of dry ice is placed in an empty 0.75 litre wire bottle tightly closed what would be the final pressure in the bottle after all CO­2 has been evaporated and temperature reaches to 25°C?

Solution

w = 20g dry CO2 which will evaporate to develop pressure

m = 44, V = 0.75 litre, P = ? T = 298K

PV = W / m RT

P × 0.75 = 20 / 44 5  0.0821 × 298

P = 14.228 atm

Pressure inside the bottle = P + atm pressure  = 14.828 + 1 = 15.828 atm

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Question 8:

The pressure of the atmosphere is 2 × 10–6 mm at about 100 mile from the earth and temperature is – 180°C. How many moles are there in 1 ml gas at this attitude?

Solution

Given, P = 2×10–6 / 760 atm

T = – 180 + 273 = 93 K

V = 1 ml = 1 / 1000 litre

PV = nRT

2 × 10–6 / 760 = n × 0.0821 × 93

n = 3.45 × 10–13 mole

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Question 9:

50 litre of dry N2 is passed through 36g of H2O at 27°C. After passage of gas, there is a loss of 1.20g in water. Calculate vapour pressure of water.

Solution

The water vapours occupies the volume of N2 gas i.e. 50 litre

∴ For H2O vapour V = 50 litre, w = 1.20g, T = 300K, m = 18

PV = w / m RT or P × 50 = 1.2 / 18 × 0.0821 × 300

∴ P = 0.03284 atm   = 24.95 mm

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Question 10:

A mixture of CO and CO2 is found to have a density of 1.50g/litre at 30°C and 730 mm. What is the composition of the mixture?

For a mixture of CO and CO2, d = 1.50 g/litre

P = 730 / 760 atm, T = 303K

PV = w / m RT; PV w / Vm  RT

730 / 760 = 1.5 / m × 0.0821 × 303 =  ∴ 38.85

i.e. molecular weight of mixture of CO and CO2 = 38.85

Let % of mole of CO be a in mixture then

Average molecular weight = a × 28 + (100 – a) 44 / 100

38.85 = 28a + 4400 – 44a / 100

a = 32.19

Mole % of CO = 32.19

Mole % of CO2 = 67.81

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Question 11:

The average speed of an ideal gas molecule at 27°C is 0.3 m sec–1. Calculate average speed at 927°C.

Solution

uav= √8RT / πM

Given uav = 0.3 m sec–1 at 300K

u1= 0.3 = √8R × 300 / πM

at T = 273 + 927 = 1200K

u2 = √8R × 1200 / πM

u2 / 0.3 = √1200 / 300 u2 = 0.6 m sec–1

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Question 12:

Pure O2 diffuses through an aperture in 224 seconds, whereas mixture of O2 and another gas containing 80% O2 diffuses from the same in 234 sec. What is the molecular weight of gas?

Solution

For gaseous mixture 80% O2, 20% gas

Average molecular weight Mm = 32 × 80 + 20 × m / 100

Now, for diffusion of gaseous mixture and pure O2

rO2 / rm = √Mm / MO2 or VO2 / EO2 × Em / Vm = √Mm / MO2

∴ 1 / 224 × 234 / 1 = √Mm / 32 ∴ Mm= 34.92    ∴ 32 × 80 + 20 × m / 100 = 34.92

∴ m = 46.6

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Question 13:

Calculate the pressure exerted by 10–23 gas molecules, each of mass 10–22 g in a container of volume one litre. The rms speed is 105 cm sec–1.

Solution

Given, n = 1023 m = 10–22g V = 1 litre = 103cc

urms = 105cm sec–1

PV = 1/3 mnu2rms

P × 103 = 1/3 × 10–22 × 1023 × (105)2

P = 3.3 × 107dyne cm–2

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Question 14:

Calculate the temperature at which CO2 has the same rms speed to that of O2 at STP.

Solution

rrms of O2 = √3RT / M  at STP, urms of O2 = √3R × 273 / 32

For CO2 urms CO2 = √3Rt / 44

Given both are same; 3R × 273 / 32 = 2RT / 44

∴ T =375.38 K = 102.38°C

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Question 15:

Calculate the compressibility factor for CO2, if one mole of it occupies 0.4 litre at 300K and 40 atm. Comment on the result.

Solution

Compressibility factor (Z) = PV / nRT

Z = 40 × 0.4 / 1 × 0.0821 × 300 = 0.65

Z value is lesser than 1 and thus, nRT > PV. In order to have Z = 1, volume of CO2must have been more at same P and T or CO2 is more compressible than ideal gas.

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