A hot air balloon has a volume of 2800 m3 at 99°C. What is the volume

1.	A hot air balloon has a volume of 2800 m3 at 99°C. What is the volume if the air cools to 80°C?
Answer» Given : V₁ = Initial volume = 2800 m³, T₁ = Initial temperature = 99°C = 99 + 273.15 = 372.15 K, T₂ = Final temperature = 80°C = 80 + 273.15 K = 353.15 K To find : V₂ = Final volume Formula : \(\frac{V_1}{T_1}\) = \(\frac{V_2}{T_2}\) (at constant n and P) Calculation : According to Charles’ law, \(\frac{V_1}{T_1}\) = \(\frac{V_2}{T_2}\) (at constant n and P) ∴ V₂ = \(\frac{V_1T2}{T_1}\) = \(\frac{2800\times 353.15}{372.15}\) = 2657 m³ ∴ The volume of the balloon when the air cools to 80°C is 2657 m³.

A hot air balloon has a volume of 2800 m3 at 99°C. What is the volume if the air cools to 80°C?

Answer»

Given :

V₁ = Initial volume = 2800 m³,

T₁ = Initial temperature = 99°C

= 99 + 273.15

= 372.15 K,

T₂ = Final temperature = 80°C

= 80 + 273.15 K

= 353.15 K

To find :

V₂ = Final volume

Formula :

\(\frac{V_1}{T_1}\) = \(\frac{V_2}{T_2}\) (at constant n and P)

Calculation :

According to Charles’ law,

\(\frac{V_1}{T_1}\) = \(\frac{V_2}{T_2}\) (at constant n and P)

∴ V₂ = \(\frac{V_1T2}{T_1}\)

= \(\frac{2800\times 353.15}{372.15}\)

= 2657 m³

∴ The volume of the balloon when the air cools to 80°C is 2657 m³.