# A Goodyear blimp typically contains 4770 m3 of helium (He) at an absolute pressure of 1.10 x 105 Pa. The temperature of the helium is 274 K. What is the mass (in kg) of the helium in the blimp?

## To find the mass of the helium in the blimp, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

First, let's convert the absolute pressure from pascals (Pa) to atmospheres (atm). Since 1 atm = 1.01325 x 10^5 Pa, the absolute pressure in atmospheres is:

1.10 x 10^5 Pa / 1.01325 x 10^5 Pa/atm ≈ 1.085 atm

Next, convert the volume from cubic meters (m^3) to liters (L). Since 1 m^3 = 1000 L, the volume in liters is:

4770 m^3 x 1000 L/m^3 = 4,770,000 L

Now, we can rearrange the ideal gas law equation to solve for the number of moles (n). The equation becomes:

n = PV / RT

Where R is the ideal gas constant, which is 0.0821 L·atm/(mol·K). Now, substitute the values into the equation:

n = (1.085 atm) x (4,770,000 L) / [(0.0821 L·atm/mol·K) x (274 K)]

n ≈ 210,029 moles

Finally, to find the mass of the helium, we can use the molar mass of helium, which is approximately 4 grams/mole.

Mass = n x molar mass
Mass = 210,029 moles x 4 g/mole = 840,116 grams

Converting grams to kilograms:

Mass = 840,116 g x (1 kg / 1000 g) ≈ 840 kg

Therefore, the mass of the helium in the blimp is approximately 840 kg.