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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 ______________________________________________ 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 ____________________________________________________ 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 __________________________________________________________ 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 _________________________________________________ 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 ________________________________________________________________________ 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 __________________________________________________ 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 _________________________________________________________ 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 __________________________________________________________ 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 _____________________________________________________________ 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 ____________________________________________________________________ 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 __________________________________________________________________________ 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 ________________________________________________________________________ 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 ____________________________________________________________________________ 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 ____________________________________________________ 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. Related Resources
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?
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
______________________________________________
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
____________________________________________________
Calculate the temperature at which 28g N2 occupies a volume of 10 litre at
2.46 atm
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
__________________________________________________________
A gas occupies 300 ml at 27°C and 730 mm pressure what would be its volume at STP
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
_________________________________________________
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.
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
________________________________________________________________________
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.
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
__________________________________________________
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?
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
_________________________________________________________
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?
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
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.
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
_____________________________________________________________
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
____________________________________________________________________
The average speed of an ideal gas molecule at 27°C is 0.3 m sec–1. Calculate average speed at 927°C.
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
__________________________________________________________________________
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?
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
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.
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
____________________________________________________________________________
Calculate the temperature at which CO2 has the same rms speed to that of O2 at STP.
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
Calculate the compressibility factor for CO2, if one mole of it occupies 0.4 litre at 300K and 40 atm. Comment on the result.
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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