Class 11 - Chemistry - States of Matter
What will the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30o C?
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
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?
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
Using the equation of state pV = nRT, show that at a given temperature density of a gas is proportional to gas pressure p.
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]
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?
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
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.
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
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?
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
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?
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
= (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
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?
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
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?
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
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?
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.
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?
Volume of air in the flask at 27o C = V cm3
(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)
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).
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
Calculate the total number of electrons present in 1.4 g of dinitrogen gas.
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
How much time would it take to distribute one Avogadro number of wheat grains, if 1010 grains are distributed each second?
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.
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.
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
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.
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
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.
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
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?
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
A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen.
Calculate the partial pressure of dihydrogen.
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
What would be the SI unit for the quantity (pV2T2/n)?
(Nm-2) (m3)2 (K2) / (mole)= Nm4 K2 mol-1
In terms of Charles’ law explain why -273o C is the lowest possible temperature.
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 −273∘C is the lowest possible temperature.
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?
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.
Explain the physical significance of van der Waals parameters?
Physical significance ‘a’ represents the magnitude of intermolecular attractive forces within gas.
Physical significance ‘b’ represents volume of gas molecule.