This chapter explains about the existence of different states of matter in terms of balance between intermolecular forces and thermal energy of particles .The chapter also explains the laws governing behaviour of ideal gases along with their application in various real life situations. It also explains the behaviour of real gases along with the conditions required for real gases. It also provides the difference between gaseous state and vapours. It also explains properties of liquids in terms of intermolecular attractions.
Volume of gas(V1) = 500dm3 (given)
Pressure of gas (p1) = 1 bar (given)
Volume of compressed gas (V2) = 200dm3 (given)
Now ,let P2 be the pressure required to compress the gas.
Therefore at constant temperature P1 V1 = P2 V2 (Boyle’s Law)
As a result P2 = P1 V1 / V2 = 1 X 500/200 = 2.5 bar
Therefore, the minimum pressure required is 2.5 bar.
According to Boyle’s Law P1 V1 = P2 V2
Here the temperature is constant. Therefore
1.2 X 120 = P2 X 180
OR
P2 = 1.2 X 120 /180 = 0.8 bar
Therefore, the pressure would be 0.8 bar.
The equation of state is given by, pV = nRT ……….. (i)
Where, p → Pressure of gas
V → Volume of gas
n→ Number of moles of gas
R → Gas constant
T → Temperature of gas
From equation (i) we have,
p = n RT/V
Where n= Mass of gas(m)/ Molar mass of gas(M)
Putting value of n in the equation, we have
p = m RT/ MV ------------(ii)
Now density(ρ) = m /V ----------------(iii)
Putting (iii) in (ii) we get
P = ρ RT / M
OR
ρ = PM / RT
Hence, at a given temperature, the density (ρ) of gas is proportional to its pressure (P)
Using the relationship of density (ρ) of the substance at temperature (T)
We have ρ = Mp/RT
or p = ρ RT/ M
For the given data if M is the molar mass of the gaseous oxide, we have
2 = ρ RT/ M ……………(1)
Also for nitrogen 5 = ρ RT/28 ………………….(2)
From (1) & (2), we have
5/2 = M/28
Or
M = 5 X 28/ 2 = 70g/mol
Hence, the molecular mass of the oxide is 70 g/mol.
Mass of gas A , WA = 1g
Mass of gas B, WB = 2g
Pressure exerted by the gas A = 2 bar
Total pressure due to both the gases = 3 bar
In this case temperature & volume remain constant
Now if MA & MB are molar masses of the gases A & B respectively,therefore
pA V= WA RT/MA & Ptotal V = (WA/MA + WB/ MB) RT
= 2 X V = 1 X RT/MA & 3 X V = (1/MA + 2/MB)RT
From these two equations, we get
3/2 = (1/MA + 2/MB) / (1 /MA) = (MB + 2MA) / MB
This result in 2MA/ MB = (3/2) -1 = ½
OR MB = 4MA
Thus, a relationship between the molecular masses of A and B is given by
4MA = MB
Aluminium reacts with caustic soda in accordance with the reaction
The reaction of aluminium with caustic soda can be represented as:
Therefore volume of hydrogen at STP released when 0.15g of Al reactsv
=0.15 x 3 x 22.4 /54 = 187ml
Now P1 = 1 bar,
P2 = 1 bar
T1 = 273 K
T2 = 20 + 273 = 293 K
V1 = 187 ml
V2 = x
When pressure is held constant,then
V2 = P1 V1 T2 / P2 T1
OR
x = 1 X 187 X 293 / 0.987 X 273
= 201 ml
Therefore, 201 mL of dihydrogen will be released.
Given,
Mass of carbon dioxide = 4.4 g
Molar mass of carbon dioxide= 44g/mol
Mass of methane = 3.2g
Molar mass of methane = 16g/mol
Now amount of methane, nCH4 = 3.2/ 16 = 0.2 mol
& amount of CO2 nCO2 = 4.4/ 44 = 0.1 mol
Also we know ,
Pv = (nCH4 + nCO2) RT
OR P X 9 = (0.2 +0.1) x 0.0821 x 300
Or p = 0.3 x 0.0821 x 300 / 9 = 0.821 atm
Hence, the total pressure exerted by the mixture is 0.821 atm
From the equation Pv = n RT for the two gases. We can write
0.8 x 0.5 = nH2 x RT or nH2 = 0.8 x 0.5 / RT
ALSO 0.7 x 2.0 = n02 . RT or n02 = 0.7 x 2 / RT
When introduced in 1 L vessel, then
Px1L = (n02 + nH2) RT
Putting the values, we get
P = 0.4 + 1.4 = 1.8 bar
Hence, the total pressure of the gaseous mixture in the vessel is 1.8 bar
For an ideal gas
Density ρ = p x M/ RT
F rom the given data,
5.46 = 2 x M/ 300 x R ………(1)
ALSO ρSTP = 1 x M/ 273 x R ………..(2)
FROM (1) & (2) ,WE GET
ρSTP = (1 x M/ 273 x R) x (300 x R/ 2 x M) x 5.46
= 300 x 5.46 / 273 X 2 = 3.00 g/dm3
Hence, the density of the gas at STP will be 3 g dm–3.
Given,
p = 0.1 bar
V = 34.05 mL = 34.05 × 10–3 L = 34.05 × 10–3 dm3
R = 0.083 bar dm3 K–1 mol–1
T = 546°C = (546 + 273) K = 819 K
From the gas equation PV = w. RT / M, we get
M = w. RT/ Pv ……….(1)
Substituting the given values in the equation (1), we get
M = (0.0625 / 0.1 x 34.04) X 82.1 X 819 = 123.46 g/mol
Hence, the molar mass of phosphorus is 123.46 g mol–1.
Let the volume of air in the flask at 27°C be V1 & that of the same amount of the gas at 477°C be V2.
According to charle’s law
V2/T2 = V1/ T1 …………………(1)
NOW volume of gas expelled out = V2 – V1, THEN
Fraction of the gas expelled out = (V2 – V1 ) / V2 = 1- (V1/ V2) …………….(2)
Also from equation (1) V1/ V2 = T1/ T2 …………..(3)
Substituting the values of (3) in (2), we get
Fraction of the air expelled = 1- T1/ T2 = (T2 – T1)/ T2
= 750- 300 /750 = 0.6
Hence, fraction of air expelled out is 0.6 or 3/5 th
Given,
Amount of the gas, n = 4.0 mol
Volume of the gas, V = 5 dm3
Pressure of the gas, p = 3.32 bar
R = 0.083 bar dm3 K–1 mol–1
From the ideal gas equation, we get
pV = nRT
⇒ T = pV / nR
= (3.32x5) / (4x0.083)
= 50K
Hence, the required temperature is 50 K.
Mass of the nitrogen = 1.4 g
Molar mass of nitrogen = 28 g mol–1
Therefore amount of nitrogen = 1.4 / 28 = 0.05 mol
Number of nitrogen molecule in 0.05 mol
= 6.023 x 1023 x 0.05
= 3.0115 x 1022
Also number of electrons in one molecule of nitrogen = 14
Therefore, total number of electrons in 1.4 g of nitrogen
= 3.0115 x 1022 x 14
= 4.22 x 1023
We know , Avogadro number = 6.023 × 1023
Therefore , time taken for distributing the grains
= (6.023 × 1023 x 1) / 1010 = 6.023 x 1013 s
By Converting these seconds into years, we get
= (6.023 x 1013) / (60 x60x24x365)
= 1,90,9800 y
Hence, the time taken would be 1.909 x 106 years.
Given,
Mass of oxygen = 8 g, molar mass of oxygen = 32 g/mol
Mass of hydrogen = 4 g, molar mass of hydrogen = 2 g/mol
Therefore amount of oxygen = 8/ 32 = 0.25 mol
And amount of hydrogen = 4/2 = 2 mol
From the gas equation PV = n RT, we get,
P X 1 = (0.25 + 2) X 0.083 X 300 = 56.02 bar
Hence, the total pressure of the mixture is 56.02 bar.
Payload of the ballon = mass of the displaced air – mass of the ballon
Radius of the ballon, r = 10 m
Mass of the ballon, m = 100kg
Therefore volume of the ballon = 4/3πr3 = 4/3 x 22/7 x (10)3 = 4190.5 m3
Now volume of the displaced air = 4190.5 m3
Given,
Density of air = 1.2 kg m–3
Therefore, the mass of the displaced air
= 4190.5 x 1.2 = 5028.6 kg
Let w be the mass of helium gas filled into the ballon,then
PV = (w/m)RT
OR w = PVM/RT
= (1.66 X 4190.5 X 103 X 4) / (0.083 X 300)
= 1117 kg (approx)
Total mass of the balloon filled with He = 1117 + 100 = 1217 kg
Therefore payload of the balloon = 5028.6 – 1217 = 3811.6 kg
Hence, the pay load of the balloon is 3811.6 kg.
From the gas equation,
PV = (w/M) RT
Where w is the mass of the gas, M is the molar mass of the gas
For CO2 , M = 44 g/mol
Substituting the given values ,we get
1 x V = (8.8 / 44) x 0.083 x 304.1
= 5.05 L
Hence, the volume occupied is 5.05 L.
From the gas equation,
PV = (w/M) RT
Substituting the given data in the gas equation, we get
PV = (2.9 / M) x R x 368
&
PV = (0.184 / 2) x R x 290
From these two equation, we can write
(2.9 / M) x R x 368 = (0.184 / 2) x R x 290
By, striking throug R from both side, we get
(2.9 / M) x 368 = (0.184 / 2) x 290
Or
(2.9 / M) = (0.092 X 290) / 368
Or
M = 2.9 x 368 / 0.092 x 290
= 40 g/mol
Hence, the molar mass of the gas is 40 g mol–1.
Pressure of the gas mixture = 1 bar
Let us consider 100g of the mixture
So ,mass of hydrogen in the mixture = 20 g
& mass of oxygen in the mixture = 80 g
Using the respective molar masses, we get
nH = 20/ 2 = 10 mol
. nO = 80 / 32 = 2.5 mol
Then, pH = XH x Ptotal
= (nH / nH + nO) x P total
= (10 / 10 + 2.5 ) x 1
= 0.8 bar
Hence, the partial pressure of dihydrogen is 0.8 bar.
The SI unit of the given quantity is obtained by substituting the SI units of all the quantities in its expression.
Also
The SI unit for pressure, p is Nm–2.
The SI unit for volume, V is m3.
The SI unit for temperature, T is K.
The SI unit for the number of moles, n is mol.
Therefore, the SI unit for quantity pV2T2 / n is given by substituting the above values in the given expression
pV2T2 / n = (Nm-2) x (m3)2 x K2 / mol
= Nm4K2 mol-1
Charles observed that the volume of certain amount of a gas changes by VO / 273.15 for each degree rise or fall in temperature.VO being the volume at 0°C.The volume at any temperature t°C is given by
Vt = VO(1 + t°C/ 273.15)
Now ,if the temperature is lowered, it becomes clear that at t = -273.15°C
Vt = VO(1+ (-273.15/273.15))
= VO = (1-1) = 0
And below t = -273.15°C, Vt becomes negative, which is not possible
Therefore the physically significant lowest temperature is -273.15° C
The maximum temperature at which a gas can be converted into a liquid by an increase in pressure is called its critical temperature(Tc). This means that the intermolecular forces of attraction between the molecules of a gas are directly proportional to its critical temperature. Hence, intermolecular forces of attraction are stronger in the case of CO2.
The van der Waals equation is an equation of state for a fluid composed of particles that have a non-zero volume and a pairwise attractive inter particle force (such as the van der Waals force). The equation is van der Waals equation
Physical significance of ‘a’: ‘a’ is a measure of the magnitude of intermolecular attractive forces between the particles
Physical significance of ‘b’:
‘b’ is the volume excluded by a mole of particles
p is the pressure of fluid
V is the total volume of container containing the fluid