A balloon is filled with a gas to a certain volume at a certain pressure at 21.9°C. If the pressure exerted on the balloon is doubled, what must the temperature (in °C) be so that the volume of the balloon doesn't change. Enter to one decimal place.

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

[tex]P\propto T[/tex]

or,

[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure = P

[tex]P_2[/tex] = final pressure = 2P

[tex]T_1[/tex] = initial temperature = [tex]21.9^oC=273+21.9=294.9K[/tex]

[tex]T_2[/tex] = final temperature = ?

Now put all the given values in the above equation, we get:

[tex]\frac{P}{294.9K}=\frac{2P}{T_2}[/tex]

[tex]T_2=589.8K[/tex]

Thus, the final temperature of balloon is, 589.8 K

andromache andromache · Answer 2 · 2022-02-11T15:14:50+01:00

The final temperature of the balloon is, 589.8 K

Given that,

A balloon is filled with a gas to a certain volume at a certain pressure at 21.9°C

Gay-Lussac's Law means the pressure of the gas i.e. directly proportional to the temperature of the gas with respect to the same volume and number of moles.

calculation of the temperature;

[tex]P \div 294.4K = 2p \div T_2\\\\T_2 = 589.8K[/tex]

Here

initial pressure = P

and, final pressure = 2P

learn more about the volume here: https://brainly.com/question/13707148