A hot air balloon is filled with 1.13 × 10 6 L 1.13×106 L of an ideal gas on a cool morning ( 11 ∘ C ) . 11 ∘C). The air is heated to 127 ∘ C . 127 ∘C. What is the volume of the air in the balloon after it is heated?Assume that none of the gas escapes from the balloon.L=_________.

Respuesta :

Answer:

New volume = [tex]1.59\times 10^6[/tex] L

Explanation:

Using Charle's law  

[tex]\frac {V_1}{T_1}=\frac {V_2}{T_2}[/tex]

Given ,  

V₁ = [tex]1.13\times 10^6[/tex] L

V₂ = ?

T₁ = 11 °C

T₂ = 127 °C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (11 + 273.15) K = 284.15 K  

T₂ = (127 + 273.15) K = 400.15 K  

Using above equation as:

[tex]\frac{1.13\times 10^6}{284.15}=\frac{V_2}{400.15}[/tex]

[tex]V_2=\frac{1.13\times 10^6\cdot \:400.15}{284.15}[/tex]

New volume = [tex]1.59\times 10^6[/tex] L