Question

The Goodyear blimps, which frequently fly over sporting events, hold approximately \(175,000 \mathrm{ft}^{3}\) of helium. If the gas is at \(23^{\circ} \mathrm{C}\) and \(1.0 \mathrm{~atm}\), what mass of helium is in the blimp?

Short answer

The mass of helium in the blimp is approximately 815041.33 grams, calculated using the Ideal Gas Law and given volume, temperature, and pressure.

Step-by-step solution

Convert temperature to Kelvin

To use the Ideal Gas Law, we first need to convert the temperature from Celsius to Kelvin. The conversion formula is: K = °C + 273.15 \( T(K) = 23^{\circ} C + 273.15 = 296.15 K \)

Write the Ideal Gas Law equation and substitute the given values

We write the Ideal Gas Law equation as PV = nRT, where P is pressure, V is volume, n is the number of moles of the gas, R is the Ideal Gas Constant, and T is temperature. Then, substitute the given values: \( (1.0 \mathrm{~atm})(175,000 \mathrm{ft^3}) = n(0.08206 \frac{\mathrm{atm} \cdot \mathrm{L}}{\mathrm{mol} \cdot \mathrm{K}})(296.15 \mathrm{K}) \)

Convert volume from cubic feet to liters

The Ideal Gas Law requires volume to be in liters, so we need to convert the given volume from cubic feet to liters. \( 1 \mathrm{ft^3} = 28.3168466 \mathrm{L} \) \( 175,000 \mathrm{ft^3} \times \frac{28.3168466 \mathrm{L}}{1 \mathrm{ft^3}} = 4955434.165 \mathrm{L} \)

Solve for the number of moles of helium

Now, we can substitute the converted volume in liters and solve the ideal gas equation for the number of moles of helium. \( (1.0 \mathrm{~atm})(4955434.165 \mathrm{L}) = n(0.08206 \frac{\mathrm{atm} \cdot \mathrm{L}}{\mathrm{mol} \cdot \mathrm{K}})(296.15 \mathrm{K}) \) \( n = \frac{(1.0 \mathrm{~atm})(4955434.165 \mathrm{L})}{(0.08206 \frac{\mathrm{atm} \cdot \mathrm{L}}{\mathrm{mol} \cdot \mathrm{K}})(296.15 \mathrm{K})} \) \( n = 203590.29 \, \mathrm{mol} \)

Calculate the mass of helium

Finally, we calculate the mass of helium by multiplying the number of moles with the molar mass of helium. The molar mass of helium is 4.0026 grams per mol. \( mass = (\mathrm{203590.29\, mol})(4.0026 \frac{\mathrm{g}}{\mathrm{mol}}) = 815041.33\, \mathrm{g} \) The mass of helium in the blimp is approximately 815041.33 grams.