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Lady_Fox [76]
3 years ago
12

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 t

o 200 atm, what is the new temperature if the volume remains constant?
(Show work pls :)!)

Chemistry
1 answer:
victus00 [196]3 years ago
6 0

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!

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