Question

for amixture of 0.5 mole of helium and 0.3 mole of N2 in a container of 0.82litre at 27°c

Answer 1

Answer:

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Answer 2

Explanation:

pressure inside the container is..

p(N2)+p(He)

partial pressure

p(T)= p(N2)+p(He)

N2

pv=nRT

R is constant 0.0821

n = 0.3 mol

t in kelvin only

k=c+273=27+273=300k

pv=nrt

p=nrt/v

p=0.3×0.0821×300/0.82 = 9.01 atm

He

n=0.5mol

t ,v, r same

p=nrt/v

p=0.5×0.0821×300/0.82= 15.01 atm

TOTAL PRESSURE =9.01+15.01=24.02atm

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