Thread: The radius decreases, How does the density of the balloon change?

1. The radius decreases, How does the density of the balloon change?

Isaac takes a round balloon out into the snow. It shrinks as the air in the balloon cools off. The radius decreases from 15cm to 14cm. How does the density of the balloon change?

I do not know how to solve this, all I can think of is density=mass/volume

Would be great if someone could help. Thanks! 2. Re: The radius decreases, How does the density of the balloon change?

the mass remains the same, only the volume changes ...

$\Delta \rho = \rho_f - \rho_i$

$\rho_f = \dfrac{m}{V_f}$

$\rho_i = \dfrac{m}{V_i}$

recall $V = \dfrac{4\pi r^3}{3}$

3. Re: The radius decreases, How does the density of the balloon change?

Suppose the mass is M grams. Originally the radius is 15 cm so the volume is $\displaystyle \frac{4}{3}\pi(225)= 4\pi(75)= 290\pi$ cubic centimeters and its density is $\displaystyle \frac{M}{290 \pi}$ grams per cubic centimeter.

The radius changes to 14 cm. What is the new volume? What is the new density?

4. Re: The radius decreases, How does the density of the balloon change?

Ok so we could call the mass anything. Because we just want to know if the density will increase or decrease.

Density is p = m / v

We find out the volume of a sphere (balloon) with:

V = 4 * pi * r 3/3

Then you can put this in p = m / v

And get p = m/(4 * pi * r^3) / 3 ---> 3m/(4 * pi * r^3)

Now we can investigate whether the density will increase or decrease. Because the mass does not matter, we can call it for 1.

P = 3 * 1 / (4 * pi * 15 ^ 3) = 7952 g / cm^3

P = 3 * 1 / (4 * pi * 14 ^ 3) = 6465 g / cm^^ 3

Do not know if I've been thinking right and I do not really know what to do now ... Because the answer should be that it increases by 23%.

5. Re: The radius decreases, How does the density of the balloon change?

If you want a percentage change in density ...

$\dfrac{\rho_f - \rho_i}{\rho_i} = \dfrac{\frac{3m}{4\pi r_f^3}-\frac{3m}{4\pi r_i^3}}{\frac{3m}{4\pi r_i^3}} = r_i^3\left(\dfrac{1}{r_f^3} -\dfrac{1}{r_i^3}\right) = \dfrac{r_i^3}{r_f^3}-1$

$\dfrac{15^3}{14^3}-1 \approx 0.23$, or a 23% increase.

6. Re: The radius decreases, How does the density of the balloon change? Originally Posted by selin Ok so we could call the mass anything. Because we just want to know if the density will increase or decrease.

Density is p = m / v

We find out the volume of a sphere (balloon) with:

V = 4 * pi * r 3/3

Then you can put this in p = m / v

And get p = m/(4 * pi * r^3) / 3 ---> 3m/(4 * pi * r^3)

Now we can investigate whether the density will increase or decrease. Because the mass does not matter, we can call it for 1.

P = 3 * 1 / (4 * pi * 15 ^ 3) = 7952 g / cm^3

P = 3 * 1 / (4 * pi * 14 ^ 3) = 6465 g / cm^^ 3
This is just bad arithmetic. The denominators are 4pi (15^3)= 42411.5 and 4pi (14^3)= 34482.1, both much larger then 3. Those fractions are much smaller than 1, not in the thousands. I have no idea how you got those values!

Do not know if I've been thinking right and I do not really know what to do now ... Because the answer should be that it increases by 23%.
Do you understand what "percentage increase" means?