Ahot air balloon filled with 1.27x10^6l of an ideal gas on a cool morning of 11 degrees c. the air is heated to 105 degrees c. what is the volume of the air in the ballon after it is heated? assume that none of the gas escapes from the balloon. l=

New volume = L

Explanation:

Using Charle's law  

Given ,  

V₁ = L

V₂ = ?

T₁ = 11 °C

T₂ = 105 °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 = (105 + 273.15) K = 378.15 K  

Using above equation as:

New volume = L

the objects have the same charge.