Question

A weather balloon is inflated to a volume of 2.2 x 10^3 L with 374 g of helium. What is the density of helium in grams per liter?

Answer 1

The density (d) of a substance is calculated by dividing the mass (m) by the given volume (v). Mathematically,

d = m / v

Substituting the known values,

d = 374 g / (2.2 x 10^3 L) = 0.17 g/L

Thus, the density of the weather balloon is 0.17 g/L.

Explanation: If the cylinders were at 1.00 atm, they would contain 50.00 L each, so at 100.0 atm they contain 5000. L each.

1m

3

=1000L,

So 100.0m

3

=1.000×10

5

L.

Since the pressure in the balloon is only 0.10 atm, though, the gas volume equals only 1.000×10

4

Lat 1 atm.

(1.000 × 10⁴ L) / (5000. L/cylinder) = 2.00 cylinders.

5000L

1.000×10

4

L

​

/cylinder =2 cylinders

Answer 2

The density (d) of a substance is calculated by dividing the mass (m) by the given volume (v). Mathematically,
d = m / v
Substituting the known values,
d = 374 g / (2.2 x 10^3 L) = 0.17 g/L
Thus, the density of the weather balloon is 0.17 g/L.