Let for gases 1, 2 and 3 the partial pressure are P₁, P₂ and P₃
Mathematically P Total = P₁ + P₂ + P₃
when we consider the gases to behave ideally
\(P_1=n_1\frac{RT}{V}; P_2=n_2 \frac{RT}{V}; R_3=n_3 \frac{RT}{V}\)
\(P_{Total}=n_1\frac{RT}{V}+n_2 \frac{RT}{V}+n_3 \frac{RT}{V}\)
\(=(n_1+n_2+n_3) \frac{RT}{V}\)
\(P_1=n_{Total}(\frac{RT}{V})\)
The partial pressure can be expressed as mole fraction multiplied X_i by the total pressure
\(P_i=X_i P_{Total}\)
\(X_i=\frac{n_i}{\sum { n_i} }\)
=\(\frac{number\ of \ mol\ of\ the\ gas 1}{Total \ number\ of\ mol of\ all \ gases}\)

The isotherms of carbon dioxide at different temperatures which is shown.in figure.
From the plots we can infer the following:

At low temperature isotherms, for example, at 13°C as the pressure increases, the volume decreases along AB and is a gas until the point B is reached. At B, a liquid separates along the line BC, both the liquid and gas co-exist and the pressure remains constant. At C, the gas is completely converted into liquid. If the pressure is higher than at C, only the liquid is compressed so, there is no significant change in the volume. The successive isotherms shows similar trend with the shorter flat region. i.e. The volume range in which the liquid and gas coexist becomes shorter. At the temperature of 31.1°C the length of the shorter portion is reduced to zero at point P. In other words, the CO₂ gas is liquefied completely at this point. This temperature is known as the liquefaction temperature or critical temperature of CO₂, At this point the pressure is 73 atm. Above this temperature CO₂ remains as a gas at all pressure values. It is then proved that many real gases behave in a similar manner to carbon dioxide.

We must calculate'R in the equation R = \(\frac{PV}{nT}\)
(R is also known as the universal gas constant)
For standard conditions in which P is 1 atm., volume 22.414 dm³. for 1 mole at273.15 (~273) K.
\(R=\frac{1 atm.\times 22.414 dm^{3}}{1 mol. \times 273.15 K}\)
= 0.0821 dm³ atm. mole^-1 K^-1
Where P = 10⁵ pascal, V = 22.71 x 10^-3 m³ for 1 mole of a gas at 273.15 K
\(R=\frac{10^{5} Pa \times 22.71 \times 10^{-3}m^{3}}{1 mol \times 273.15 K}\)
= 8.314 Pa m³ K^-1 mol-1
= 8.314 x 10^-2 bar dm³ K^-1 mol^-1
= 8.314 J K^-1 mol^-1

PV = nRT. Here T and V are constant.
Hence \(\frac{P_1}{P_2}=\frac{n_1}{n_2}=\frac{w_1}{w_2}\)
Where P₁ and P₂ are the initial and final pressures.
i.e., P₁= 10 atm; P₂ = 8 atm.
w₁ and w₂ weight of the gas present initiates and after leaking,
w₁ = 15 kg : w₂ = ?
\(\therefore \frac{10}{8}= \frac{15}{w_2} (or) w_2=12 kg\).
Hence, the amount of gas leaked is 15 - 12 = 3 kg.

b = 4V (or) V = \(\frac{b}{4}=\frac{0.0318}{4}\) = 7.95 x 10^-3L mol^-1 = 7.95 cm³ mol^-1
Volume occupied by one mole of O₂ molecule
\(=\frac{7.95}{6.02\times 10^{23}}=1.32 \times 10^{-23} cm^{3}\)
Considering the molecule to be spherical, \(V=\frac{4}{3} \pi r^{3}=1.32 \times 10^{-23}cm^{3}\)
\(r^{3}=3.15\times 10^{-24}\)
Solving r = \(1.466 \times 10^{-8}\)cm
Diameter 2r = 2 x 1.466 x 10^-8 cm = 2.932 x 10^-8 cm
= 2.932 \(\overset {0}{A}\)