If the pressure of a real gas O_{2} in a cont | vedclass

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(B) The given equation is $P = \frac{RT}{2V - b} - \frac{a}{4b^{2}}$.Rearranging the terms, we get: $P + \frac{a}{4b^{2}} = \frac{RT}{2V - b}$.Factor

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$16$

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(B) The given equation is $P = \frac{RT}{2V - b} - \frac{a}{4b^{2}}$.Rearranging the terms, we get: $P + \frac{a}{4b^{2}} = \frac{RT}{2V - b}$.Factor out $2$ from the denominator: $P + \frac{a}{4b^{2}} = \frac{RT}{2(V - b/2)}$.Multiplying both sides by $2(V - b/2)$, we get: $(P + \frac{a}{4b^{2}})(V - b/2) = \frac{RT}{2}$.Comparing this with the van der Waals equation for $n$ moles: $(P + \frac{an^{2}}{V^{2}})(V - nb) = nRT$, we can see that for the given equation to hold, $n = 1/2 	ext{ mole}$.The molar mass of $O_{2}$ is $32 	ext{ g/mol}$.Therefore, the mass of the gas is $m = n 	imes M = (1/2) 	imes 32 	ext{ g} = 16 	ext{ g}$.

If the pressure of a real gas $O_{2}$ in a container is given by $P = \frac{RT}{2V - b} - \frac{a}{4b^{2}}$, then the mass of the gas in the container is: (in $\text{ g}$)

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