Question

A hot-air balloon is filled with air to a volume of at 750. torr and 21°C. The air in the balloon is then heated to 58°C, causing the balloon to expand to a volume of . What is the ratio of the number of moles of air in the heated balloon to the original number of moles of air in the balloon? (Hint: Openings in the balloon allow air to flow in and out. Thus the pressure in the balloon is always the same as that of the atmosphere.) Ratio = 1

Answer 1

`(n_2)/(n_1)=0.93`

Step-by-step explanation:

Hello, in this case the the volume at both the first and second states are missing, therefore, one could suppose a typical value around 4000 m³ for the first state and 4200 m³ based on their normal operating conditions. In such a way, by considering the normal gas law given in terms of the change of temperature, moles and volume, assuming constant pressure as said on the statement as follows:

`(n_1T_1)/(V_1)= (n_2T_2)/(V_2)`

One solves for the increasing mole ratio as shown below:

`(n_2)/(n_1)=(T_1V_2)/(T_2V_1)=((294.15K)(4200m^3))/((331.15K)(4000m^3))=0.93`

Thus, the volume at the second state is less than that of the first state.

Best regards.

Answer 2

The ratio of the number of moles in the heated balloon to the original number of moles is less than 1.

Step-by-step explanation:

When the air of the hot-air balloon is heated, its density decreases. It means that the molecules inside the balloon (at 58°C) occupy a bigger volume than the same amount of molecules at the beginning of the process (which are at 21°C). Due to the fact that the balloon envelope is flexible but not infinitely, part of the molecules tha were originally inside the balloon must flow out in order to release some additional space inside the balloon for the remaining molecules. This extra space makes possible a relatively constant difference of densities between the air inside and the air outside, which gives buoyancy to the balloon. And this is why the amount of moles in the balloon is slightly lower after heating than it was originally.