Question

The air inside a hot-air balloon is heated to 59.00 °C (Tinitial) and then cools to 39.00 °C (Tfinal). By what percentage does the volume of the balloon change?

Answer 1

6.1%

Step-by-step explanation:

Assuming pressure inside balloon remains constant during the temperature change.

Therefore, as per Charles' law at constant pressure,

`(V_1)/(T_1) =(V_2)/(T_2)`

`T_2=39\°\ C=39+273=312\ K`

`T_1=59°\ C=273+59=332\ K`

`V_2=V_1* (T_2)/(T_1) \\=V_1*(312)/(332)\\=0.939V_1`

Percentage change in volume

`\%\ change\ in\ volume=(V_1-0.939V_1)/(V_1) * 100\\=6.1\%`

Change in volume of the balloon is 6.1%