Question

What will be the temperature difference needed in a hot air balloon to lift $1.00 \,kg$ of mass? Assume that the volume of the balloon is $100 m ^{3}$, the temperature of the ambient air is $298\, K ,$ the pressure is 1.00 bar, and the air is an ideal gas with average molar mass of $29 \,g \,mol ^{-1}$.

Answer 1

For a hot air balloon, the lifting capacity, $\Delta m$ is determined by substracting the mass of hot air from the mass of cold air that is displaced by balloon.

$\Delta m=m$ (cold) $-m$ (hot)

$=[$ moles (cold) $-$ moles (hot)] $\times$ molar mass

$=\left[\frac{p V}{R T(\text { cold })}-\frac{p V}{R T( hot )}\right] \times$ molar mass

$\Delta m=1 \,kg =1000 \,g ,$

molar mass $=29 \,g \,mol ^{-1}$

$\therefore \frac{1000}{29}=\frac{p V}{R}\left[\frac{1}{T(\text { cold })}-\frac{1}{T(\text { hot })}\right]$

$p=1.0\, bar =10 \times 10^{5}$ pascal

$V=100 \,m ^{3}$

$R=8.314\, m ^{3} \,Pa\, K ^{-1} mol ^{-1}$

$\left[\frac{1}{T(\text { cold })}-\frac{1}{T( hot )}\right]$

$=\frac{1000 g \times 8.314 \,m ^{3} PaK ^{-1} mol ^{-1}}{29 \,g\, mol ^{-1} \times 1.0 \times 10^{5} Pa }$

$\left[\frac{1}{298}-\frac{1}{T( hot )}\right]=2.87 \times 10^{-5} \times 100 m ^{3}$

$\frac{1}{T( hot )}=\left[\frac{1}{298}-2.87 \times 10^{-5}\right]$

$=3.33 \times 10^{-3} K ^{-1}$

$\therefore T( hot )=300\, K$

Difference $=(300-298)=2\, K$