Ahot air balloon is filled with 1.37 × 106 l of an ideal gas on a cool morning (11 °c. the air is heated to 117 °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.

So according to Charles's law the volume of a gas is directly proportional to the absorbed absolute temperature assuming constant pressure and a closed system. Or V/T=constant. 
(V1/T1)=(V2/T2). Absolute temperature is in degrees Kelvin so:
 
T1=273.15+11= 284.15 K and T2=273.15 + 117=390.15 K. V1= 1.37*10^6 L.

We need to find V2. V2= (V1/T1)*T2.

Now we plug in the numbers: V2=(1.37*10^6/284.15)*390.15=1881068.1 L.
The volume of the air balloon when we heat the gas inside is
V2=1.88*10^6 L. 

We first assume the gas is an ideal gas so we can use the equation PV = nRT. At constant pressure, nR/P are constants so at any condition the ratio of V/T is the same. With this, we calculate as follows:

V1/T1 = k = V2/T2
1.37 × 10^6 L / (11+273.15) = V2 / (117+273.15)
V2 = 1881068.1 L

Ithink the answer would be 1,350!