Question

A gas in a sealed container has a pressure of 125 atm at a temperature of 303 K. If the pressure in the container is increased to 200 atm, what is the new temperature if the volume remains constant?
(Show work pls :)!)

Answer 1

Answer:

$T_2=484.8K$

Explanation:

Hello there!

In this case, according to the the variable temperature and pressure and constant volume, it turns out possible for us to calculate the new temperature via the Gay-Lussac's law as shown below:

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

Thus, by solving for the final temperature, T2, we obtain:

$T_2 =\frac{T_1P_2}{P_1}$

So we plug in the given data to obtain:

$T_2 =\frac{303K*200atm}{125atm}\\\\T_2=484.8K$

Best regards!