Question #777e2

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Question #777e2

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Answer 1 · 2016-05-12T06:27:11

The pressure is unknown. The partial pressure is a multiple of the mole fraction of each component.

Explanation:

We have volume, and amount of substance; we do not have temperature, which we require if pressure is to be calculated. I assume that the temperature is $298 \cdot K$ $298*K$ .

$Moles of helium:$ $"Moles of helium:"$ $\frac{32.6 \cdot g}{4.0 \cdot g \cdot m o l^{- 1}} = 8 \cdot m o l$ $(32.6*g)/(4.0*g*mol^-1)=8*mol$

$Moles of dioxygen:$ $"Moles of dioxygen:"$ $\frac{51.2 \cdot g}{32.0 \cdot g \cdot m o l^{- 1}} = 1.6 \cdot m o l$ $(51.2*g)/(32.0*g*mol^-1)=1.6*mol$

The total pressure $P = \frac{n R T}{V}$ $P=(nRT)/V$ $=$ $=$

$\frac{(8.0 + 1.6) \cdot m o l \times 0.0821 \cdot L \cdot a t m \cdot K^{- 1} \cdot m o l^{- 1} \times 298 \cdot K}{0.150 \cdot L}$ $((8.0+1.6)*molxx0.0821*L*atm*K^-1*mol^-1xx298*K)/(0.150*L)$ $=$ $=$ $P_{T}$ $P_T$

The partial pressure of dioxygen $P_{O_{2}} = \frac{P_{O_{2}}}{P_{T}}$ $P_(O_2)=(P_(O_2))/P_T$ $=$ $=$ $\frac{1.6 \cdot m o l}{(8 + 1.6) \cdot m o l} \times P_{T}$ $(1.6*mol)/((8+1.6)*mol)xxP_T$ .

This sounds like a deep sea diving mix, where helium replaces dinitrogen in order to avoid the narcotic effect of dinitrogen.

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Question #777e2

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