Question

A gaseous mixture of an equal mass of ${\mathrm{CH}}_4$ and ${\mathrm{O}}_2$ is stored over water. if the total pressure is $0.5\mathrm{atm}$ then the partial pressure of dry ${\mathrm{CH}}_4$ gas is ?(aqueous tension $=0.2\mathrm{atm}$)

Answer 1

Step 1 : Given data:

A gaseous mixture contains an equal mass of ${\mathrm{CH}}_4$ and ${\mathrm{O}}_2$

Molecular Mass of ${\mathrm{CH}}_4=16$

Molecular Mass of ${\mathrm{O}}_2=32$

Step 2: Calculation of total moles :
Moles of ${\mathrm{CH}}_4=\frac{\mathrm{x}}{16}$
Moles of ${\mathrm{O}}_2=\frac{\mathrm{x}}{32}$

Hence the total moles $=\left(\frac{\mathrm{x}}{16}\right)+\left(\frac{\mathrm{x}}{32}\right)=\frac{3\mathrm{x}}{32}$

Step 3: Calculation of Mole fractions of ${\mathrm{CH}}_4$:

Mole fractions of CH₄ can be calculated as:

$$\begin{gathered} =\frac{\mathrm{moles}\mathrm{of}{\mathrm{CH}}_4}{\mathrm{total}\mathrm{moles}} \\ =\frac{\left({\displaystyle \frac{\mathrm{x}}{16}}\right)}{\left(\frac{3\mathrm{x}}{32}\right)} \\ =\frac{2}{3} \end{gathered}$$

Step 4:Formula for calculation of partial pressure of any gas:

According to Dalton’s law of partial pressure, the total pressure of a mixture of gases is equal to the sum of the partial pressures of the individual gases in the mixture. Dalton’s law is perfectly true for the ideal gas mixture.

$P_{total}=P_1+P_2+P_3+......P_i$

Where P_total = total pressure of the ideal gas mixture

${\mathrm{P}}_1,{\mathrm{P}}_2,{\mathrm{P}}_3,{\mathrm{P}}_{\mathrm{i}}$are the partial pressures of the gases $1,2,3\dots \mathrm{i}$respectively

In a mixture of gases, the pressure basically exerted by an individual gas is defined as its partial pressure.

The partial pressure of an individual gas of the ideal gas mixture can be given as:

$$P_i$$=$x_i\times p$

Where, $$P_i$$= Partial pressure

$x_i$= mole fraction

$p\mathit{=}$total pressure

Step 5: Calculation of partial pressure of ${\mathrm{CH}}_4$:

The partial pressure of ${\mathrm{CH}}_4$ can be calculated as:

$$P_i$$=$x_i\times p$
$$\begin{gathered} =\frac{2}{3}\mathrm{x}0.5 \\ =0.33\mathrm{atm} \end{gathered}$$

Hence, the partial pressure of dry ${\mathrm{CH}}_4$ gas is calculated to be $0.33\mathrm{atm}.$