Collision in which there is no loss of kinetic energy but there is transfer of energy, is called elastic collision.

If units of p = atm, units of  V = dm³  ,units of n = mol then,units of
\(a=\frac { { pV }^{ 2 } }{ { n }^{ 2 } } =\frac { atm.({ dm }^{ 3 })^{ 2 } }{ (mol)^{ 2 } } =atm \ dm^{ 6 } \ mol^{ -2 }\)

\(\left( P+\frac { { an }^{ 2 } }{ { V }^{ 2 } } \right) (V-nb)=nRT\)
\(a=\left( \frac { nRT }{ V-nb } -p \right) \times \frac { { V }^{ 2 } }{ { n }^{ 2 } }\)
\( n=2 mo,V=5 \ L,p=9.1 atm,T=273+27=300 K\)
\(R=0.082L\ atm \ { mol }^{ -1 }K^{ -1 }\)
\(\\ a=\left( \frac { 2\times 0.0082\times 300 }{ (5-2\times 0.052) } -9.1 \right) \times \frac { { (5) }^{ 2 } }{ { (2) }^{ 2 } }\)
\(=\frac { 49.2 }{ 4.896 } -9.1\times \frac { 25 }{ 4 } =(10.05\times 9.1)\times \frac { 25 }{ 4 }\)
\( =5.94 \ atm \ L^{ 2 } \ mol^{ -2 }\)

As the temperature of a liquid increases, kinetic energy of the molecules increases which can overcome intermolecular forces. So, the liquid can flow more easily, this results in decrease in viscosity of the liquid.

T = 1513.33K

Radius of the balloon = 10 m
\(\therefore\)Volume of the balloon = \(\frac{4}{3}\pi\)r² = \(\frac{4}{3}\times\frac{22}{7}\times\)(10m)³ = 4190.5 m³
Volume of He filled at 1.66 bar and 27°C = 4190.5 m3
Calculation of mass of He
PV = nRT = \(\frac { w }{ M } \)RT
or \(w\) = \(\frac{MPV}{RT}\) = \(\frac { (4\times { 10 }^{ -3 }kgmol^{ -1 })(1.66bar)(4190.5\times 103dm^{ 3 }) }{ (0.083bardm^{ 3 }K^{ -1 }mol^{ -1 })(300K) } \)
= 1117.5 kg
Total mass of the balloon along with He = 100 + 1117.5 = 1217.5 kg
Maximum mass of the 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
\(\therefore\) Pay load = 5028.6 - 1217.5 kg = 3811.1 kg

Ideal Gas: A gas that follows Boyle's law, Charles' law and Avogadro law strictly, is called an ideal gas. It is assumed that intermolecular forces are not present between the molecules of an ideal gas.
Real Gases: Gases which deviate from ideal gas behaviour are known as real gases. NH₃ is expected to show more deviation. Since NH₃ is polar in nature and it can be liquified easily.

\(\left( p+\frac { { n }^{ 2 }a }{ { V }^{ 2 } } \right) \)(V - nb) = nRT
When n is the no. of moles present and' a' and 'b' are known as van der Waals constants.
constants.
Significance of van der Waals constants
van der Waals constant ' a': 'a' is related to the magnitude of the attractive forces among the molecules of a particular gas. Greater the value of 'a', more will be the attractive forces.
Unit of 'a' = L² mol^-1
van der Waals Constant 'b': 'b' determines the volume occupied by the gas molecules
which depends upon size of molecule.
Unit of 'b' = L mol^-1.

A liquid boils when vapour pressure becomes equal to atmospheric pressure. At hill station, atmospheric pressure is low. Therefore, liquid boils at a lower temperature and cooking is not perfect. In a pressure cooker, the pressure inside the cooker increases and the liquid boils at a higher temperature.

Isochore is plot of p vs T for a definite amount of a gas at constant volume. It is a graphical representation of Gay-Lussac or Amonton's law. For different isochores at different volumes, draw a line parallel to the temperature axis representing a constant pressure and cutting the three isochores at T₁, T₂, T₃ respectively. From the graph we find that T₁ > T₂ > T₃. As V \(\propto \) T at constant p.
Thus,     V₁ > V₂ > V₃
          

At high pressure, the real gas shows large deviations from ideal behaviour as the curves are far apart.

The atmosphere pressure is less at hill station due to which water boils below 100\(^{0}\)C and vegetables and pulses take ling time in cooking

From ideal gas equation we have,
\(pV=nRT \ or \ p=\frac { m }{ v } \times \frac { RT }{ M } =\frac { dRT }{ M } \)
\( d \ =\frac { pM }{ RT } \)
\( or \ d\alpha \frac { p }{ T } \ [R \ and \ M \ constant \ for \ a \ given \ gas]\)
\( \frac { { d }_{ 1 } }{ { d }_{ 2 } } =\frac { { p }_{ 1 } }{ { p }_{ 2 } } \times \frac { { T }_{ 2 } }{ { T }_{ 1 } } or \ { d }_{ 2 }=\frac { { d }_{ 1 }\times { p }_{ 2 }\times { T }_{ 1 } }{ { p }_{ 1 }\times { T }_{ 2 } }\)
\( { d }_{ 1 }=6.46 \ g \ { dm }^{ -3 },\ { p }_{ 1 }=5bar,\)
\({ T }_{ 1 }=273+25=298k\)
\( { d }_{ 2 }=?,\ { p }_{ 2 }=1 \ bar,{ T }_{ 2 }=27k\)
\( { d }_{ 2 }=\frac { 6.46g{ dm }^{ -3 }\times 1 \ bar\times 298k }{ 5 \ bar\times 273k }\)
\( =1.41 \ g{ dm }^{ -3 }\)

P₁ = 760 mm Hg, V₁ = 600 mL
T₁ = 25 + 273 = 298 K
V₂ = 640 mL and T₂=10 + 273 = 283 K
According to combined gas law,
\(\frac { { P }_{ 1 }V_{ 1 } }{ { T }_{ 1 } } =\frac { { P }_{ 2 }V_{ 2 } }{ { T }_{ 2 } } \)
\(\Rightarrow \)  P₂ = \(\frac { { p }_{ 1 }{ V }_{ 1 }{ T }_{ 2 } }{ { T }_{ 1 }{ V }_{ 2 } } \)
\(\Rightarrow \) P₂ = \(\frac { (760 \ mm \ Hg)\times (600 \ mL)\times (283K) }{ (640mL)\times (298K) } \)
= 676.6 mm Hg

From the available data:
V₁ = 25 m³  T₁ = 27 + 273 = 300K
V₂ = ?   T₂ = 77 + 273 = 350K
Since the pressure of the gas is constant, Charles' law is applicable
\(\frac { { V }_{ 1 } }{ { V }_{ 2 } } =\frac { { T }_{ 1 } }{ T_{ 2 } } \) or V₂=\(\frac { { V }_{ 1 }\times { V }_{ 2 } }{ { T }_{ 1 } } \)
v₂ = \(\frac { \left( { 25m }^{ 3 } \right) \times \left( 350 \right) K }{ (300K) } \) = 29.17 m³.

 Density, d = pM/RT
For same gas at different temperatures and pressures
\(\cfrac { { d }_{ 2 } }{ d_{ 1 } } =\cfrac { { p }_{ 2 }{ T }_{ 1 } }{ { p }_{ 1 }{ T }_{ 2 } }\)
\( { d }_{ 2 }=\cfrac { { p }_{ 2 }{ T }_{ 1 }d_{ 1 } }{ { p }_{ 1 }{ T }_{ 2 } } \)
\(\cfrac { 1\times 300\times 5.46 }{ 2\times 273 } \)
= 3 g dm^-3

80.25 mol^-1

Compound = SO₂ Mass of one molecule of gas = 1.07 x 10^-22g

Temperature by ideal gas equation = 602.4 K
Temperature by van der waals equations = 153.38 k