Question

A balloon under 14.7 atm of pressure at 273K gets left in a car in Florida and the temperature increases to 303K. What is the new pressure in the balloon?
a. 44.7 atm
b. 13.2 atm
c. 16.3 atm
d. 5627 atm​

Answer 1

Answer:

The new pressure in the balloon is 16.3 atm

Explanation:

Gay-Lussac's law establishes the relationship between the temperature and the pressure of a gas when the volume is constant.

This law establishes that pressure and temperature are directly proportional quantities, that is to say that if the temperature increases the pressure increases and if the temperature decreases the pressure decreases.

Mathematically, Gay-Lussac's law states that, when a gas undergoes a constant volume transformation, the quotient of the pressure exerted by the gas temperature remains constant:

$\frac{P}{T}=k$

When you want to study two different states, an initial one and a final one, of a gas, you can use the expression:

$\frac{P1}{T1}=\frac{P2}{T2}$

In this case:

• P1= 14.7 atm
• T1= 273 K
• P2= ?
• T2= 303 K

Replacing:

$\frac{14.7 atm}{273 K}=\frac{P2}{303 K}$

Solving:

$P2=303K*\frac{14.7 atm}{273 K}$

P2= 16.3 atm

The new pressure in the balloon is 16.3 atm