In a reaction involving the collection of gas by downward displacement of water, the pressure of dry vapor collected can be calculated using Dalton's law.
P_{dry gas collected} = P_Total - P_{water vapour}
P_{water vapour} is generally referred as aqueous tension and its values are available for air at various temperatures. These values are relevant in weather forecast.

\(Z=\frac { { PV }_{ real } }{ nRT } \) ......(1)
\({ V }_{ ideal }=\frac { nRT }{ P } \)  ......(2)
Substituting (2) in (1)
\(Z=\frac { { V }_{ real } }{ { V }_{ ideal } } \)
Where V_real is the molar volume of the real gas and ideal is the molar volume of it when it behaves ideal. i.e. compressibility is the ratio of real volume of the. gas to ideal volume of any gas.

(i) Boyle's point is the temperature at which, real gases behave ideally over a range of low pressures.
(ii) The Boyle point varies with the nature of the gas.
(iii) Above the Boyle point, Z > 1 for real gases. i.e., they show positive deviation.
(iv) Below the Boyle point, the real gases first show a decrease for Z, reaches a minimum and then increase with increase in pressure.

The volume of a gas increases with temperature at constant pressure. Each line (isobar) represents the variation of volume with temperature at certain pressure. The pressure increases from PI to P. i.e. P₁ < P₂ < P₃ < P₄ < P₅ When these lines are extrapolated or extended to zero volume, they' intersect at a temperature of - 273.15°C. All the gases are becoming liquids if they are cooled to sufficiently low temperatures. In other words, all gases occupy zero volume at absolute zero

According to Boyle 's law for a given mass of a gas under two different sets of conditions at constant temperature we can write
P₁V₁ = P₂V₂ = k           . ...(1)

\(\frac { { \gamma }_{ unknown } }{ \gamma { N }_{ 2 } } =\frac { { t }_{ { N }_{ 2 } } }{ { t }_{ unknown } } =\sqrt { \frac { { m }_{ { N }_{ 2 } } }{ { m }_{ unknown } } } \)
\(\frac { 84\quad sec }{ 192\quad sec } =\sqrt { \frac { 14g\quad { mol }^{ -1 } }{ { m }_{ unknown } } } =\frac { 14g\quad { mol }^{ -1 } }{ { m }_{ unknown } } \)
\({ m }_{ unknown }=28 g { mol }^{ -1 }\times { \left( \frac { 192\quad sec }{ 84\quad sec } \right) }^{ 2 }\)
m_unknown = 146 g mol^-1 . 

Given,
V = 154.4 \(\times\) 10^-3 dm³,
P = 742 mm of Hg
T = 298 K m = ?
\(n=\frac { PV }{ RT } =\frac { 742mm\ Hg\times 154.4\times { 10 }^{ -3 }L }{ 62mm\ Hg\ L{ K }^{ -1 }{ mol }^{ -1 }\times 298K } \)
\(n=\frac { Mass }{ Molar\ Mass } \)
mass = n \(\times\) Molar mass
= 0.006 \(\times\) 2.016
= 0.0121 g = 12.1 mg.

V₁ = 8.5 dm³ V₂ = 6.37 dm³
T₁ = ? T₂ = 0^o C = 273 K
\(\frac { { V }_{ 1 } }{ { T }_{ 1 } } =\frac { { V }_{ 2 } }{ { T }_{ 2 } } \)
\({ V }_{ 1 }\times \left( \frac { { T }_{ 2 } }{ { V }_{ 2 } } \right) ={ T }_{ 1 }\)

T₁ = 364.28 K

T₁ = 15^oC + 273 T₂= 38 + 273
T₁ = 288 K  T₂ = 311 K
V₁ = 2.58 dm³ V₂ = ?
(P = 1 atm constant)
\(\frac { { V }_{ 1 } }{ { T }_{ 1 } } =\frac { { V }_{ 2 } }{ { T }_{ 2 } } \)
\({ V }_{ 2 }=\left( \frac { { V }_{ 1 } }{ { T }_{ 1 } } \right) \times { T }_{ 2 }\)

V₂ = 2.78 dm³ i.e. volume increased from 2.58 dm³ to 2.78 dm³.

n_A = 1.5 mol n_B = ?
V_A= 37.6 dm³ V_B = 16.5 dm³
(T = 298 K constant)
\(\frac { { V }_{ A } }{ { n }_{ A } } =\frac { { V }_{ B } }{ { n }_{ B } } \)
\({ n }_{ A }=\left( \frac { { n }_{ A } }{ { V }_{ A } } \right) { V }_{ B }\)

= 0.66 mol.