Question

If the ratio of masses of $S{{O}_{3}}$ and ${{O}_{2}}$ gases confined in a vessel is 1:1 then the ratio of their partial pressure would be:
(A) 5:2
(B) 2:5
(C) 2:1
(D) 1:2

Answer 1

This question will have the use of ideal gas law that state a relation between pressure (P), volume (V) and Temperature (T)
For an ideal gas,
$PV=nRT$
Where, P is pressure, V is volume, T is temperature, n is the quantity of gas (expressed in moles) and R is the gas constant.

Complete Solution :
Let the mass of $S{{O}_{3}}$ and ${{O}_{2}}$ enclosed in the vessel be M.
Therefore number of $S{{O}_{3}}=\dfrac{M}{80}$
Similarly, number of moles of ${{O}_{2}}=\dfrac{M}{32}$
Now, partial pressure of $S{{O}_{3}}$ (denoted by${{P}_{A}}$)
$\Rightarrow {{P}_{A}}=\dfrac{M}{80}\left( \dfrac{R\times T}{V} \right)$………………………………………….(i)

Similarly,
Partial pressure of ${{O}_{2}}$ (denoted by${{P}_{B}}$)
$\Rightarrow {{P}_{B}}=\dfrac{M}{32}\left( \dfrac{R\times T}{V} \right)$…………………………………………..(ii)
On dividing equation (i) and (ii)

Now, $\dfrac{{{P}_{A}}}{{{P}_{B}}}=\dfrac{m}{80}\times \dfrac{32}{m}=\dfrac{2}{5}$
Therefore, the ratio of partial pressure of and will be $2:5$
So, the correct answer is “Option B”.

Note: The total pressure of a mixture of gases equals the sum of the partial pressures of all the gases in the mixture. Partial pressure helps us in predicting the movement of gases. This is because gases tend to equalize their pressure in two regions that are connected. Since, each gas in a mixture behaves independently of the other gases, we can therefore use the ideal gas law to calculate the partial pressure of each gas in the mixture.