Assuming the molar mass of air is 28.84 , g , | vedclass

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(A) The payload is the difference between the mass of air displaced by the balloon and the mass of $He$ gas inside the balloon. Using the ideal gas eq

Vedclass · Accepted Answer

$1.01 	imes 10^3 \, kg$

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(A) The payload is the difference between the mass of air displaced by the balloon and the mass of $He$ gas inside the balloon. Using the ideal gas equation $PV = nRT = (m/M)RT$,the mass $m = (PVM)/(RT)$. Given: $P = 1 \, atm$,$V = 10^6 \, L$,$T = 27 + 273 = 300 \, K$,$R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1}$. Mass of air displaced $(m_{air})$ = $(1 	imes 10^6 	imes 28.84) / (0.0821 	imes 300) \approx 1170.36 \, kg$. Mass of $He$ inside $(m_{He})$ = $(1 	imes 10^6 	imes 4) / (0.0821 	imes 300) \approx 162.32 \, kg$. Payload = $m_{air} - m_{He} = 1170.36 - 162.32 = 1008.04 \, kg \approx 1010 \, kg$.

Assuming the molar mass of air is $28.84 \, g \, mol^{-1}$,calculate the payload (lifting capacity) of a balloon of volume $10^6 \, L$ filled with $He$ at $27 \, ^\circ C$ and $1 \, atm$ pressure.

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