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
To do this, we make use of the following representations
So. we have the following derivative
dP = dV - P * dV
Divide through by dV
dP/dV = 1 - P
Next, we set up the integral
[tex]\int\limits^x_{0.040} \frac {1}{1 - P}dP= \int\limits^5_0 {dV}[/tex]
Integrate both sides
[tex]-\ln(|P - 1|)|\limits^x_{0.04} = 5[/tex]
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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