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_{i} =x_{i}*P

x_{i} = n_{i}/n

n(O_{2}) = 1.42 g / 32 g/mol = 0.044 mol

n(N_{2}) = 1.00 g / 28 g/mol = 0.0357 mol

n(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 = 0.033

x_{i}(N2) = 0.0357/1.33 = 0.027

x_{i} (Ar) = 1.25 / 1.33 =0.94

P_{i}(O2) = 0.033*1.5= 0.0495 atm

P_{i}(N2) = 0.027 * 1.5 = 0.0405 atm

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

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

Learn more about our help with Assignments: Inorganic Chemistry

## Comments

## Leave a comment