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Answer :
Answer:
123.5 kPa
Explanation:
P2=P1T2/T1
You can check this by knowing that P and T at constant V have a proportional relationship. Hence, this is correct.
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Final answer:
According to Gay-Lussac's law, the pressure in the sealed container will increase to approximately 123.4 kPa when the temperature increases to 376 K.
Explanation:
The question is about the relationship between gas temperature and pressure in a closed container, which is established by Gay-Lussac's law in physics. Gay-Lussac's law states that the pressure of a gas is directly proportional to its temperature if the volume and the number of moles of the gas are kept constant. In mathematical terms, P1/T1 = P2/T2, where P1 and T1 are the initial pressure and temperature, and P2 and T2 are the final pressure and temperature.
Let's use this formula to solve your question. We are given: P1 = 97.9 kPa, T1 = 298 K, and T2 = 376 K. We need to find P2. Rearrange the formula to solve for P2: P2 = P1 * (T2/T1) = 97.9 kPa * (376 K / 298 K) = 123.4 kPa.
Therefore, the pressure inside the container when the temperature rises to 376 K is approximately 123.4 kPa.
Learn more about Gay-Lussac's Law here:
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