Question

What is the total pressure after all he stopcocks are opened at oom temperature?

Answer 1

Answer:

Supposing the invisible diagram shows connections among all three flasks, so that when all the stopcocks are opened, there is one contiguous, larger volume for the gases to occupy:

207 torr / (760 torr/atm) x 1.15 L = 0.31322 L·atm He

0.456 atm x (1.15 L) = 0.5244 L·atm Ne

(24.5 kPa) / (101.325 kPa/atm) x (2.45 L) = 0.59240 L·atm Ar

0.31322 L·atm He + 0.5244 L·atm Ne + 0.59240 L·atm Ar = 1.43002 L·atm total

b)

(1.43002 L·atm) / (1.15L + 1.15L + 2.45 L) = 0.301 atm total after all gases mix

a)

(0.31322 L·atm He) / (1.43002 L·atm total) x (0.301 atm total) = 0.0659 atm He

(0.5244 L·atm Ne) / (1.43002 L·atm total) x (0.301 atm total) = 0.110 atm Ne

(0.59240 L·atm Ar) / (1.43002 L·atm total) x (0.301 atm total) = 0.125 atm Ar

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