Question

A hot air balloon is filled with 1.31 × 10 6 L of an ideal gas on a cool morning ( 11 ∘ C ) . The air is heated to 121 ∘ 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.

Answer 1

Answer:

$1.82\times 10^6 L$ is the volume of the air in the balloon after it is heated.

Explanation:

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$ (at constant pressure)

where,

$V_1\text{ and }T_1$ are the initial volume and temperature of the gas.

$V_2\text{ and }T_2$ are the final volume and temperature of the gas.

We are given:

$V_1= 1.31\times 10^6 L\\T_1=11^oC=(11+273.15)K=284.15K\\V_2=?\\T_2=121^oC=(121+273.15)K=394.15 K$

Putting values in above equation, we get:

$\frac{1.31\times 10^6 L}{284.15 K}=\frac{V_2}{394.14 K}\\\\V_2=\frac{V_1\times T_2}{T_1}$

$V_2=1.82\times 10^6 L$

$1.82\times 10^6 L$ is the volume of the air in the balloon after it is heated.