Question

A hot air balloon is filled with 1.41 à 106 l of an ideal gas on a cool morning (11 °c). the air is heated to 109 °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 : The volume of the air in the balloon after it heated is, $1.896\times 10^6L$

Solution : Given,

Initial volume of air balloon = $1.41\times 10^6L$

Initial temperature of air balloon = $11^oC=273+11=284K$   $(0^oC=273K)$

Final temperature of air balloon = $109^oC=273+109=382K$

According to the Charles' law, the volume of an ideal gas is directly proportional to the temperature of the gas at constant pressure.

It is represented as,

$V\propto T$

or, $\frac{V_1}{V_2}=\frac{T_1}{T_2}\\V_2=\frac{T_2}{T_1}\times V_1$

where,

$V_1$ = initial volume of air balloon

$V_2$ = final volume of air balloon

$T_1$ = initial temperature of air balloon

$T_2$ = final temperature of air balloon

Now put all the given values in the above formula, we get the final volume of air balloon.

$V_2=\frac{382K}{284K}\times (1.41\times 10^6L)=1.896\times 10^6L$

Therefore, the volume of the air in the balloon after it heated is, $1.896\times 10^6L$