The mass percentage composition of dry air at | vedclass

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Question

(B) To find the partial pressure, we first calculate the mole fraction of each component based on the given mass percentages. Assume $100 \ g$ of air.

Vedclass · Accepted Answer

$0.75 p \ atm, 0.17 p \ atm, 0.08 p \ atm$

Vedclass · Answer

(B) To find the partial pressure, we first calculate the mole fraction of each component based on the given mass percentages. Assume $100 \ g$ of air. Moles of $N_2 = \frac{63 \ g}{28 \ g/mol} = 2.25 \ mol$. Moles of $O_2 = \frac{16 \ g}{32 \ g/mol} = 0.5 \ mol$. Moles of $Kr = \frac{21 \ g}{84 \ g/mol} = 0.25 \ mol$. Total moles $= 2.25 + 0.5 + 0.25 = 3.0 \ mol$. Partial pressure of a gas is given by $p_i = x_i 	imes p_{total}$, where $x_i$ is the mole fraction. $p_{N_2} = (\frac{2.25}{3.0}) p = 0.75 p \ atm$. $p_{O_2} = (\frac{0.5}{3.0}) p = 0.167 p \ atm$. $p_{Kr} = (\frac{0.25}{3.0}) p = 0.083 p \ atm$. Comparing these values with the given options, option $(B)$ is the closest approximation.

The mass percentage composition of dry air at sea level is approximately $N_2: 63 \%$, $O_2: 16 \%$, $Kr: 21 \%$. If the total pressure is $p \ atm$, calculate the partial pressure of each component respectively. (Molecular weights: $N_2 = 28, O_2 = 32, Kr = 84$)

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