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?
You asked this twice. See my response to your later post.
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.