A balloon with volume 2800 m³ is heated up to 60 °C. If the outer temperature is 12 °C and the air has a pressure of 960 hPa, what is the density of the air inside the balloon and on the outside?

In the solutions they used the following formula: $ \rho_1 = \rho_0 \cdot \frac{P_1}{P_0} \cdot \frac{T_0}{T_1} $

However, I have no clue on how to get to this fomula. It is neither in my fomulary nor was I able to derive it from another formula.

My approach was to calculate the molar mass of the air which is $28.96 \frac{g}{mol}$ and then use it in this formula (which I derived from the $p\cdot V = n \cdot R \cdot T $): $$ \rho = \frac{p\cdot M}{R\cdot T} $$ But that doesn't seem to be correct, since I get extremly high numbers and the correct value should be $1.17 \frac{kg}{m^3}$. Also, this approach only works for the outer density - I do not have the pressure inside the baloon.


1 Answer 1


$$\rho=\frac{n*M}{V}$$ With $n$ the number of moles of the gas, $M$ the molar mass of the gas, and $V$ the volume of the gas. $n$ and $M$ are constant, so: $$\rho_1=\rho_0*\frac{V_0}{V_1}$$ For $V$ we have: $$V=\frac{n r T}{p}$$ $n$ and $R$ are constant, so: $$V_1=V_0*\frac{T_1}{T_0}*\frac{p_0}{p_1}$$ So: $$\rho_1=\rho_0*\frac{T_0}{T_1}*\frac{p_1}{p_0}$$ Which is exactly the expression given in the solution.


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