Why does the density of gas in a hot air balloon decrease as the gas is heated?

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Why does the density of gas in a hot air balloon decrease as the gas is heated?

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Answer 1 · 2017-06-24T16:32:50

The density $d$ $d$ of a gas is related by the equation

$d = \frac{M P}{R T}$ $d = (MP)/(RT)$

where

$M$ $M$ is the molar mass of the gas (in $g/mol$ $"g/mol"$ )
$P$ $P$ is the pressure of the gas (in $atm$ $"atm"$ )
$R$ $R$ is the universal gas constant, equal to $0.082057 \frac{L \cdot atm}{mol \cdot K}$ $0.082057("L"·"atm")/("mol"·"K")$
$T$ $T$ is the absolute temperature of the gas (in $K$ $"K"$ )

We can see from this relationship that the density ( $d$ $d$ ) of a gas and its temperature ( $T$ $T$ ) are inversely proportional. That is, if the gas in the balloon is heated (temperature increased), its density will decrease (assuming relatively constant pressure).

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Why does the density of gas in a hot air balloon decrease as the gas is heated?

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