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,

$\frac{P_1}{P_2}=\frac{T_1}{T_2}$

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:

$\frac{x}{10x}=\frac{T_1}{T_2}$

$\frac{T_1}{T_2}=\frac{1}{10}$

$T_2=10T_1$

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.