Write a nuclear reaction for the neutron-induced fission of u−235 to form xe−144 and sr−90.
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Answer:
1. Total pressure is 475 torr.
2. The partial pressure of CO is 23.8 torr.
3. The mole fraction of CO is 0.0501.
Explanation:
We have 3 gases in different bulbs. Once the stopcocks are opened, they share the same final volume which is the sum of all individual volumes.
V = Va + Vb + Vc = 150 mL + 300 mL + 750 mL = 1200 mL
Since we know initial pressures and volumes for each gas, we can find the final pressures using Boyle's Law. The mathematical expression is
P₁ . V₁ = P₂ . V₂
We assume that temperature remains constant and that gases behave as ideal gases.
CO
P₁ = 190 torr; V₁ = 150 mL; P₂ = ?; V₂ = 1200 mL
P₁ . V₁ = P₂ . V₂
190 torr . 150 mL = P₂ . 1200 mL
P₂ = 23.8 torr
Ar
P₁ = 0.500 atm; V₁ = 300 mL; P₂ = ?; V₂ = 1200 mL
P₁ . V₁ = P₂ . V₂
0.500 atm . 300 mL = P₂ . 1200 mL
P₂ = 0.125 atm
Kr
P₁ = 75.994 kPa ; V₁ = 750 mL; P₂ = ?; V₂ = 1200 mL
P₁ . V₁ = P₂ . V₂
75.994 kPa . 750 mL = P₂ . 1200 mL
P₂ = 47.5 kPa
The total pressure is the sum of partial pressures.
P = P(CO) + P(Ar) + P(Kr) = 23.8 torr + 95.0 torr + 356 torr = 475 torr
We can find the mole fraction of of CO using the following expression, based on Dalton's Law:
P(CO) = P . X(CO)
where,
X(CO) is the mole fraction of CO
Then,
X(CO) = P(CO)/P = 23.8 torr / 475 torr = 0.0501.