### Video Transcript

A gas flows smoothly through a
pipe. The pipe’s cross-sectional area
contracts from 0.075 meters squared to 0.025 meters squared. The gas enters the pipe moving at
1.8 meters per second and leaves the pipe moving at 2.0 meters per second. The density of the gas as it enters
the pipe is 1.4 kilograms per meter cubed. What is the ratio of the density of
the gas where it enters the pipe to its density where it exits the pipe?

We’ll start by drawing a diagram of
this scenario. We have a pipe that contracts from
one cross-sectional area that we will call 𝐴 one to a second cross-sectional area
that we will call 𝐴 two. Gas flows smoothly through this
pipe and enters the pipe with a velocity that we will call 𝑣 one. The gas leaves the pipe at a
velocity that we will call 𝑣 two. Before the contraction, the gas has
a density that we will call 𝜌 one. After the contraction, the gas has
a density that we will call 𝜌 two.

The question asks us to calculate
the ratio of the density of the gas where it enters the pipe to its density where it
exits the pipe. So the question wants us to
calculate 𝜌 one divided by 𝜌 two. In the question, we are told that
the pipe’s cross-sectional area contracts from 0.075 meters squared to 0.025 meters
squared. This means that 𝐴 one is equal to
0.075 meters squared and 𝐴 two is equal to 0.025 meters squared. The question also tells us that the
gas enters the pipe moving at 1.8 meters per second and leaves the pipe moving at
2.0 meters per second. So 𝑣 one is equal to 1.8 meters
per second and 𝑣 two is equal to 2.0 meters per second. Finally, the question tells us that
the density of the gas as it enters the pipe is 1.4 kilograms per meter cubed. So 𝜌 one is equal to 1.4 kilograms
per meter cubed.

Before we go any further, it’s
important to notice that all the values given to us are in SI units. This means that we don’t have to
convert any of them before we substitute them into any equations or use them for any
calculations.

To answer this question, we’re
going to use the continuity equation for fluids, which states that a fluid’s density
multiplied by the cross-sectional area of the pipe it is flowing in multiplied by
the fluid’s velocity is constant. This means that we can write
density multiplied by cross-sectional area multiplied by velocity before contraction
is equal to density multiplied by cross-sectional area multiplied by velocity after
contraction. And we would like to rearrange this
to get an expression for 𝜌 one divided by 𝜌 two.

We will start by dividing both
sides of the equation by 𝜌 two. And we see that these 𝜌 twos on
the right cancel. Next, we will divide both sides of
the equation by 𝐴 one multiplied by 𝑣 one. And we see that these 𝐴 ones
cancel. And also these 𝑣 ones cancel on
the left. And this gives us our expression
for 𝜌 one divided by 𝜌 two, which is equal to 𝐴 two multiplied by 𝑣 two divided
by 𝐴 one multiplied by 𝑣 one. And we actually see that we don’t
need the value of 𝜌 one to calculate 𝜌 one divided by 𝜌 two.

We can now continue and substitute
our values for 𝐴 one, 𝑣 one, 𝐴 two, and 𝑣 two into this equation, which gives us
𝜌 one divided by 𝜌 two is equal to 0.025 meters squared multiplied by 2.0 meters
per second divided by 0.075 meters squared multiplied by 1.8 meters per second. Evaluating this expression gives us
𝜌 one divided by 𝜌 two is equal to 0.37 to two decimal places.

The ratio of the density of the gas
where it enters the pipe to its density where it exits the pipe is 0.37.