Question #bf572

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Question #bf572

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Answer 1 · 2014-12-11T09:58:37

The answer is $1.79 a t m$ $1.79atm$ .

This problem can be solved by using the concept of partial pressures; according to Dalton's law the final pressure in the cylinder will be equal to the sum of the partial pressures of the two gases.

So, in order to determine the final pressure, one must first calculate $A r$ $Ar$ 's partial pressure. This can be done by using the ideal gas law, $P V = n R T$ $PV = nRT$ ; we know that

$n_{A r} = \frac{m a s s}{m o l a r m a s s} = \frac{2.53 g}{40 \frac{g}{m o l}} = 0.06$ $n_(Ar) = (mass)/(molarmass) = (2.53g)/(40g/(mol)) = 0.06$ moles

From $P V = n R T \to P = \frac{n R T}{V}$ $PV = nRT -> P = (nRT)/V$ , we get

$P_{A r} = \frac{0.06 \cdot 0.082 \cdot (273.15 + 27)}{2.0} = 0.74$ $P_(Ar) = (0.06 * 0.082 * (273.15+27))/2.0 = 0.74$ atm

Therefore, $P_{T O T A L} = P_{H e} + P_{A r} = 1.05 + 0.74 = 1.79$ $P_(TOTAL) = P_(He) + P_(Ar) = 1.05 + 0.74 = 1.79$ atm

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Question #bf572

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