Ask Question
27 October, 19:48

The temperature of a sample gas in a steel tank 30.00 kpa increased from 10.0 c to 25.0c what is the final pressure inside the tank?

+4
Answers (1)
  1. 27 October, 23:10
    0
    31.59 kpa

    Explanation:

    The variables in this problem are temperature and pressure. Both variables are related by the gay lusaacs law.

    Gay-Lussac's Law: The Pressure Temperature Law. This law states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. This means that as the pressure increases so does the temperature.

    From the law, we have;

    P1/P2 = T1/T2

    From the problem;

    P1 = Intial pressure = 30 kpa

    P2 = Final pressure = x

    T1 = Initial temperature = 10 + 273 = 283K (Upon converting to kelvin temperature)

    T2 = Final temperature = 25 + 273 = 298K (Upon converting to kelvin temperature)

    Substituting into the equation, we have;

    30 / P2 = 283 / 298

    Upon cross multiplication, we have;

    P2 = (298 * 30) / 283

    P2 = 31.59 kpa
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “The temperature of a sample gas in a steel tank 30.00 kpa increased from 10.0 c to 25.0c what is the final pressure inside the tank? ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers