Answer to Question #71828 in General Chemistry for Olivia Blee

Question

A sample of 2.90 g of SO2(g) originally in a 5.20-L vessel at 26 ∘C is transferred to a 12.0-L vessel at 25 ∘C. A sample of 2.35 g N2(g) originally in a 2.50-L vessel at 20 ∘C is transferred to this same 12.0-L vessel.
What is the partial pressure of SO2(g) in the larger container?

Answer 1

p1V1/ p2 V2 = T1 /T2 
p2(SO2)= (p1(SO2)* V1(SO2)* T2(SO2))/(T1(SO2)* V2(SO2)) = (1000*n(SO2)*R* T2(SO2))/(T1(SO2)*V2(SO2)) =
=(1000*n(SO2)*R* T2(SO2))/(T1(SO2)*V2(SO2)) = (1000*0,045 mol*8,31 J/mol⋅K* 298K)/(299K*0,012m3) = 31058 Pa
n(SO2) = m(SO2)/ M(SO2) = 2,90g/ 64g/mol = 0,045 mol
p2(N2)= (p1(N2)* V1(N2)* T2(N2))/(T1(N2)*V2(N2)) = (1000*n(N2)*R* T2(N2))/(T1(N2)*V2(N2)) =
= (1000*0,084mol*8,31 J/mol⋅K* 298K)/(293K*0,012m3) = 59162 Pa
n(N2) = m(N2)/ M(N2) = 2,35g/ 28g/mol = 0,084 mol
ptot = p2(SO2) + p2(N2) =31058 Pa + 59162 Pa = 90220 Pa
p(SO2) = n(SO2)/(n(SO2)+ n(N2) ) * ptot = 0.045mol/(0.045mol +0.084mol ) *90220Pa = 31472 Pa.