Question

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

Answer 1

Answer: 2

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

$=\left[\frac{pV}{RT(\text{ cold })}-\frac{pV}{RT(hot)}\right]\times$ molar mass

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

molar mass $=29gmo{l}^{-1}$

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

$p=1.0bar=10\times 10^5$ pascal

$V=100m^3$

$R=8.314m^{3}Pa{K}^{-1}mo{l}^{-1}$

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

$=\frac{1000g\times 8.314m^{3}Pa{K}^{-1}mo{l}^{-1}}{29gmo{l}^{-1}\times 1.0\times 10^{5}Pa}$

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

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

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

$\therefore T(hot)=300K$

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