If a balloon containing 3000 L of gas at 39°C and 99 kPa rises to an altitude where the pressure is 45.5 kPa and the temperature is 16°C,
the volume of the balloon under these new conditions would be calculated using the following conversion factor ratios:
( the one I have marked, I’m not sure if that’s the answer or not )
Answers
[tex]\tt =3000~L\times \dfrac{289}{312}\times \dfrac{99}{45.5}[/tex]
Further explanationGiven
3000 L of gas at 39°C and 99 kPa to 45.5 kPa and 16°C,
Required
the new volume
Solution
Combined with Boyle's law and Gay Lussac's law
[tex]\tt \dfrac{P_1.V_1}{T_1}=\dfrac{P_2.V_2}{T_2}[/tex]
T₁ = 39 + 273 = 312
T₂ = 16 + 273 = 289
Input the value :
V₂ = (P₁V₁.T₂)/(P₂.T₁)
V₂ = (99 x 3000 x 289)/(45.5 x 312)
or we can write it as:
V₂ = 3000 L x (289/312) x (99/45.5)
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