Question

Oxygen flows through one tube into a liter flask filled with air, and the mixture of oxygen and air (considered well stirred) escapes through another tube. Assuming that air contains 4% oxygen, what percentage of oxygen will the flask contain after 5 L have passed through the intake tube?

Answer 1

The amount of oxygen in the flask through the intake tube illustrates proportions

The flask 99.35% will contain of oxygen after 5L have passed through the intake tube

How to calculate the percentage of oxygen?

To do this, we make use of the following representations

• P for the proportion of oxygen
• V for te volume of oxygen
• x for the proportion of oxygen in the flask after 5 L passed through

So. we have the following derivative

dP = dV - P * dV

Divide through by dV

dP/dV  = 1 - P

Next, we set up the integral

$\int\limits^x_{0.040} \frac {1}{1 - P}dP= \int\limits^5_0 {dV}$

Integrate both sides

$-\ln(|P - 1|)|\limits^x_{0.04} = 5$

Expand

-ln(|x - 1|) + ln(|0.04 - 1|) = 5

Evaluate the difference

-ln(|x - 1|) + ln(|-0.96|) = 5

Express as a fraction

ln(0.96/|x - 1|) = 5

Express both sides as exponents with a base of 2

e^ln(0.96/|x - 1|) = e^5

Solve for x

x = 1 - 0.96/e^5

Evaluate the difference

x = 0.9935

Express as percentage

x = 99.35%

Hence, the flask 99.35% will contain of oxygen after 5L have passed through the intake tube

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