A trash compactor can compress its contents to 0.350 times their original volume. Neglecting the mass of air expelled, by what factor is the density of the rubbish increased?
If we call 𝑣 sub 𝑐 the volume of the compressed trash, then we’re told that that volume is equal to 0.350 times the original volume which we’ll call 𝑣 sub 𝑢, for uncompressed. Keeping these subscripts, 𝑐 for compressed and 𝑢 for uncompressed, what we want to solve for is the ratio of the density of the compressed trash to the density of the uncompressed trash.
To figure this out, let’s recall a relationship between density, mass, and volume. The density 𝜌 of an object is defined as its mass divided by its volume. So in our case, 𝜌 sub 𝑐, the density of the compressed trash, equals the trash’s mass 𝑚 divided by the compressed volume, 𝑣 sub 𝑐. And the density of the uncompressed trash, 𝜌 sub 𝑢, equals the trash’s mass divided by its uncompressed volume. If we divide 𝜌 sub 𝑐 by 𝜌 sub 𝑢, if we divide the equation for 𝜌 sub 𝑐 by the equation for 𝜌 sub 𝑢, then we see the ratio we’re looking for is equal to the trash’s mass divided by its compressed volume divided by its mass divided by its uncompressed volume.
Know that the mass of the rubbish doesn’t change, so the 𝑚 is constant and can cancel out of this ratio. We can make a substitution for the compressed volume of the trash based on our given information. We know that 𝑣 sub 𝑐 is equal to 0.350 times 𝑣 sub 𝑢. We see then that the term in the denominator of each fraction, 𝑣 sub 𝑢, is common and can cancel out. So overall, our ratio of 𝜌 sub 𝑐 to 𝜌 sub 𝑢 simplifies to one divided by 0.350.
When we enter this value into our calculator, we find a ratio of 2.86. That’s how many times more dense the compacted trash was compared to the uncompacted trash.