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Mashutka [201]
3 years ago
5

The dirigible Hindenburg had 3.7E6 m3 of hydrogen in its gas bags at 1.1 atm and 7°C. What was the weight of the hydrogen in pou

nds?
Chemistry
1 answer:
zepelin [54]3 years ago
4 0
Assuming that the gas is ideal, we can use the ideal gas equation PV=nRT to calculate for the number of moles. Then, multiply the molar mass of the gas to obtain the mass. We do as follows:

PV = nRT
n = PV / RT
n = 1.1 atm (3.7x10^9 L) / 0.08205 L-atm/mol-K (280.15) = 177061931.3 mol H2

Mass = (177061931.3 mol H2) 18.02 g/mol ( 1 kg / 1000g) ( 2.2 lb / 1 kg ) = 7019443.21 lb H2
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zimovet [89]

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\large \boxed{\text{54 $\, \%$ faster }}

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\begin{array}{rcl}\dfrac{v_{2}}{v_{1}} & = & \sqrt{\dfrac{M_{1}}{M_{2}}}\\\\& = & \sqrt{\dfrac{70.91}{30.07}}\\\\& = & \sqrt{2.358}\\\\& = & \mathbf{1.54}\\\end{array}\\\text{Ethane molecules travel at 1.54 times the speed of chlorine molecules}\\\text{or $\large \boxed{\textbf{54 $\%$ faster }}$ than chlorine molecules}

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