How many litres of mixture is released each time ? please help?

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let;

x = liters of mixture that is released every time

y = liters of nitrogen that is replaced every time

but y = x ( … an equal amount of nitrogen is added )

➤

50 - x = Amount remaining in the cylinder the first time a few liters is released;

0.25(50 - x) = (12.5 - 0.25x) oxygen left

0.75(50 - x) = (37.5 - 0.75x) nitrogen left

➤

(50 - x) + x = 50 adding nitrogen first time

(37.5 - 0.75x) + x = (37.5 + 0.25x) new nitrogen amount

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(12.5 - 0.25x) = oxygen in the cylinder; 1/50 (12.5 - 0.25x)

(37.5 + 0.25x) = nitrogen in the cylinder; 1/50(37.5 + 0.25x)

➤

50 - x = Amount remaining in the cylinder the second time a few liters is released;

1/50 (12.5 - 0.25x) (50 - x) fraction of oxygen left

1/50 (37.5 + 0.25x) (50 - x) fraction of nitrogen left

➤

adding nitrogen second time:

Oxygen = 16%

Nitrogen = (100 - 16)% = 84%

1/50 (37.5 + 0.25x) (50 - x) + x = 84/100 (50)

-0.005x² + 0.5x - 4.5 = 0

x² - 100x + 900 = 0

(x - 10)(x - 90) = 0

x = 10 or x = 90 (Ignore value x = 90 since it will make 50 - 90 negative liters)

x = 10 liters

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To see how calculations work out trial an easy volume like 10 L released

50 L composed of: Oxygen 12.5 L, Nitrogen 37.5 L

If 10 L released the proportions will stay the same

40 L composed of: Oxygen 10 L, Nitrogen 30 L,

Then 10 L added with only nitrogen increasing

50 L composed of: Oxygen 10 L, Nitrogen 40 L (20% oxygen)

Again 10 L released so proportions will stay the same

40 L composed of: Oxygen 8 L, Nitrogen 32 L (same proportions)

Then 10 L added with only nitrogen increasing

50 L composed of: Oxygen 8 L, Nitrogen 42 L (20% oxygen)

The oxygen content has become 16% of the total volume

Well what do you know. (I cheated of course).

70336.1422 exactly. Every time. Unless it's a Wednesday of course when the top falls off the cheese.