Question

you place a balloon in a closed chamber at STP. You increase the chamber pressure by a factor of 10. What happens to the balloon?

Answer 1

pops because of increasing the atmospheric pressure

Answer 2

Answer : The temperature of balloon increases to 10 times of original temperature.

Explanation :

In this problem, volume remain constant. The temperature and pressure will vary.

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.

`P\propto T`

or,

`(P_1)/(P_2)=(T_1)/(T_2)`

where,

`P_1` = let initial pressure = x

`P_2` = final pressure = 10x

`T_1` = initial temperature

`T_2` = final temperature

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

`(x)/(10x)=(T_1)/(T_2)`

`(T_1)/(T_2)=(1)/(10)`

`T_2=10T_1`

From this we conclude that, the temperature increases to 10 times of original temperature.

Hence, the temperature of balloon increases to 10 times of original temperature.