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PV = nRT (1.15*10^5 Pa)(10^-3 m^3) = 3/208.5(1+a)mol * 8.314 J mol^-1 K^-1 * ...
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The answer is correct. A is the unknown mass in the problem and the unit of this is in grams (g). There should be a mixture of a particular gas and the question is to determine amount of gas in the certain container. In case you need it, here is a step procedure of the problem:
`PV = nRT`
`(1.15x10^5 Pa)*(10^-3 m^3) = (3)/(208.5 (g)/(mol)) (1+a) * (8.314 (J)/(mol*K))*(523K)`
`115 Pa-m^3 (or J or N-m) = (1+a)* ((3)(8.314(J)/(mol*K))(523 K))/(208.5 (g)/(mol))`
`115 J = (1+a)* 62.564 J/g`
`(115J)/(62.564 J/g) = 1+a`
`1.838 g = 1+a`
`a = 1.838 - 1`
`a = 0.838 grams`
Posted by jerichorayel on June 14, 2013 at 4:31 PM (Answer #1)
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