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Unformatted text preview: y ρ = m / V and the definition of molar mass, M = m / n. This leads to : PM = ρ RT or M = ρ RT / P Using SI units we get: M = ρ RT / P 6
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M = (8.161 mg/L) x (1 kg/10 mg) x (1000 L/m ) x 8.3145 J/K.mol x (273.15 + 27.85 K) / (0.100 atm x 101325 Pa/atm) Since all quantities are expressed in SI units, the result will also be in SI unit (kg/mol for molar mass) M = 0.00202 kg/mol This can only be the molar mass of molecular Hydrogen H2(g) 4 Question 10: (10 points) An ideal gas mixture contains equal masses of components A and B. Given that the molar mass of A is twice that of B and that the total pressure is PT = 0.300 bar, calculate PA, the partial pressure of the component A. Let us consider a mass mA of A and a mass mB of B. We know that: mA = mB . We are told that the molar mass of A is twice that of B. So, we write: MA = 2MB . The problem has to do with partial pressures and mixtures of ideal gases. Hence, we will need to use Dalton’s law, which for A is written as: PA = xAPT where xA is the mole fraction of A in the mixture and PT is the total pressure. Since we know PT, we can obtain the partial pressure of A if we know the mole fraction of A. This mole fraction can be derived from the two relationships involving masses and molar masses. Indeed, we can write the mass of A as the product of molar mass of A by the number of moles of A. Similarly for B. mA = nA x MA and mB = nB x MB Since mA = mB , we can write: nA x MA = nB x MB Then, using MA = 2MB , we conclude that nA x 2MB = nB x MB or 2nA = nB The mole fraction of A is given by: xA = nA / (nA + nB) = nA / (nA + 2nA) = 1/3 xA = 1/3 and PT = 0.300 bar Hence, PA = 0.100 bar Question 11: (10 points) A non ideal gas is described using the following equation of state: RT bT − a P=
+
2
Vm
Vm
 3 where a = 1.00 S.I. unit and b = 3.33 10 S.I. unit are specific parameters proposed for this gas. Part a. (5 points) Determine the Boyle’s temperature TB predicted by this equation of state for this gas and indicate 3
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whether at the state defined by T1 = 200. K and Vm,1 = 0.00500 m...
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This note was uploaded on 01/26/2014 for the course CHEM 3615 taught by Professor Aresker during the Spring '07 term at Virginia Tech.
 Spring '07
 AREsker
 Physical chemistry, pH

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