At the sea level, the dry air mass percentage composition is given as nitrogen gas : 70.0, oxygen gas : 27.0…

At the sea level, the dry air mass percentage composition is given as nitrogen gas : 70.0, oxygen gas : 27.0 and argon gas : 3.0. If total pressure is 1.15 atm , then calculate the ratio of followings respectively :
(i) partial pressure of nitrogen gas to partial pressure of oxygen gas
(ii) partial pressure of oxygen gas to partial pressure of argon gas
(Given : Molar mass of $\mathrm{N}, \mathrm{O}$ and Ar are 14, 16, and $40 \mathrm{~g} \mathrm{~mol}^{-1}$ respectively)
  1. $4.26,19.3$
  2. $2.59,11.85$
  3. $5.46,17.8$
  4. $2.96,11.2$

Solution

$\frac{\mathrm{P}_{\mathrm{N}_2}}{\mathrm{Po}_2}=\frac{\mathrm{x}_{\mathrm{N}_2} \cdot \mathrm{P}_{\mathrm{T}}}{\mathrm{x}_{\mathrm{O}_2} \cdot \mathrm{P}_{\mathrm{T}}}=\frac{\mathrm{n}_{\mathrm{N}_2}}{\mathrm{n}_{\mathrm{O}_2}}$ {using Dalton's law of partial pressure}
$\begin{aligned} & =\frac{70 / 28}{27 / 32}=2.96 \\ & \frac{\mathrm{P}_{\mathrm{O}_2}}{\mathrm{P}_{\mathrm{Ar}}}=\frac{\mathrm{n}_{\mathrm{O}_2}}{\mathrm{n}_{\mathrm{Ar}}}=\frac{27 / 32}{3 / 40}=11.25\end{aligned}$

Asked in: JEE Main 2025 (07 Apr Shift 1)

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