# Answer to Question #19872 in Inorganic Chemistry for Brittany

Question #19872

A mix of O, N, and Ar has total pressure of 1.5 atm. The weights of the gases in the mix are 1.42g, 1.00g, and .50g. Calculate partial pressure of each gas; if volume of mix is 2L, what's the temp of the gas? At this temp, what's the root mean square speed of the N molecules?

Expert's answer

P

x

n(O

n(N

n(Ar) = 50 g / 40 g/mol = 1.25 mol

x

x

x

P

P

P (Ar) = 0.94*1.5 = 1.41 atm

PV = nRT

T = PV/nR

T = 15450 Pa * 0.002 m3 / 1.33 mol *8.314 = 2.79 K

_{i}=x_{i}*Px

_{i}= n_{i}/nn(O

_{2}) = 1.42 g / 32 g/mol = 0.044 moln(N

_{2}) = 1.00 g / 28 g/mol = 0.0357 moln(Ar) = 50 g / 40 g/mol = 1.25 mol

x

_{i}(O_{2}) = 0.044/(0.044 + 0.0357+1.25) = 0.044/1.33 = 0.033x

_{i}(N2) = 0.0357/1.33 = 0.027x

_{i}(Ar) = 1.25 / 1.33 =0.94P

_{i}(O2) = 0.033*1.5= 0.0495 atmP

_{i}(N2) = 0.027 * 1.5 = 0.0405 atmP (Ar) = 0.94*1.5 = 1.41 atm

PV = nRT

T = PV/nR

T = 15450 Pa * 0.002 m3 / 1.33 mol *8.314 = 2.79 K

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