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* 10^6L`

Solution : Given,

Initial volume of air balloon =
`1.41* 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,
`(V_1)/(V_2)=(T_1)/(T_2)\\V_2=(T_2)/(T_1)* 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=(382K)/(284K)* (1.41* 10^6L)=1.896* 10^6L`

Therefore, the volume of the air in the balloon after it heated is,
`1.896* 10^6L`