Since a hot-air balloon is filled with heated air, it means that the pressure of the air inside the balloon is lower than the surrounding air. So then why doesn't the atmospheric pressure crush the balloon?
2007-09-11 11:14:17 · 2 answers · asked by DichloroDiphenyl 5 in Science & Mathematics ➔ Physics
Re: Alexander
I'm not really a physics student, so I don't know what the R and the T stand for!!
(I'm assuming V is Volume and P is pressure)
at least in meteorological terms, we hear of low pressure fronts being created when there is hot air. Could you throw light on that?
2007-09-11 11:35:22 · update #1
No, it doesn't mean lower pressure.
The pressure is the same.
But higher temperature means larger volume, and lower density.
V = RT/P
2007-09-11 11:26:15 · answer #1 · answered by Alexander 6 · 1⤊ 0⤋
No, it does not mean "...that the pressure of the air inside the balloon is lower than the surrounding air."
Hot air balloons are typically open at the bottom,
so the pressure is the same as outside.
Because the air inside is hotter, its DENSITY is
lower.
2007-09-11 18:28:55 · answer #2 · answered by Irv S 7 · 0⤊ 0⤋