Question

chemistry homework help - letter BHot air is less dense than cold air and this si the reason that hot air balloons float. A hot air balloon on a cool summer morning is filled with air to a volume of 2.60x 10^6L and warmed to 195 degrees fahrenheit(90 degrees celsius). A hot air balloon must be inflated to a volume of approximately 2.80x10^6L in order to float in air. a. At constant pressure, provide a molecular level explanation of the relationship between temperature and volume of a gas.B. Calculate the temperature the gas must be in(celsius in order to float).

Answer 1

Answer: the volume air necessary must be at 117.93 °C to float

Step-by-step explanation:

The question requires us to determine the final temperature of air in order to float in a hot air baloon, given its initial temperature and volume and the volume necessary for the baloon to float.

The following information was provided by the question:

initial volume = V1 = 2.60x 10^6 L

initial temperature = T1 = 90 °C

final volume = V2 = 2.80x10^6 L

Considering that the pressure remains constant, we can apply Charle's Law, which matematically states the relationship between temperature and volume of a gas at constant pressure:

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

Rearranging the equation to calculate the final temperature, we'll have:

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

Note that the temperature must be used in Kelvin, as it refers to the absolute temperature of the gas. We can convert the initial temperature given from °C to K as:

`\begin{gathered} T(K)=T(°C)+273.15 \\ T(K)=90°C+273.15=363.15K \end{gathered}`

And, applying the values provided by the question:

`T_2=((2.80*10^6L)*(363.15K))/((2.60*10^6L))=391.08K`

And we can conver this temperature from K to °C as:

`\begin{gathered} T(K)=T(°C)+273.15\rightarrow T(°C)=T(K)-273.15 \\ T(°C)=391.08-273.15=117.93°C \end{gathered}`

Therefore, the volume air necessary must be at 117.93 °C to float.