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?

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

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