A hot air balloon is filled with 1.39 × 106 L of an ideal gas on a cool morning (11 °C). The air is heated to 105 °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.
If P is constant, then (V1/T1) = (V2/T2)
T must be in kelvin.
Is the answer 1.85x10^6 L?
yes.
How did u get it
To find the volume of the air in the balloon after it is heated, we can use the ideal gas law equation:
PV = nRT
Where:
P = Pressure
V = Volume
n = Number of moles of gas
R = Ideal gas constant
T = Temperature in Kelvin
First, let's convert the temperatures from Celsius to Kelvin. We add 273.15 to each temperature to get the following:
Initial temperature (Ti) = 11 °C + 273.15 = 284.15 K
Final temperature (Tf) = 105 °C + 273.15 = 378.15 K
The ideal gas law equation can be rearranged to solve for the final volume (Vf):
Vf = Vi * (Tf / Ti)
Where Vi is the initial volume.
Now, let's plug in the values:
Vi = 1.39 × 10^6 L
Ti = 284.15 K
Tf = 378.15 K
Vf = 1.39 × 10^6 L * (378.15 K / 284.15 K)
Calculate the final volume:
Vf = 1.85 × 10^6 L
Therefore, the volume of the air in the balloon after it is heated is approximately 1.85 × 10^6 L.