Question

For a mixture of 0.5 mole of helium and 0.3 mole of N_(2) in a container of 0.82 litre at 27°C​

Answer 1

Explanation:

v=0.82l

t in k =c +273=27+273=300k

partial pressure of N2=

n=0.3

r=0.0821

pv=nrt,,p=nrt/v=0.3×0.0821×300/0.82=9.01 atm

n=0.5

partial pressure of He

r=0.0321

pv=nrt

p=nrt/v

=0.0821×0.5×300/0.82 = 15.01 atm

TOTAL PRESSURE EQUAL TO SUM OF THE

PARTIAL PRESSURES

P( T )=P(N2)+P(He) =15.01+9.01=24.02atm