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When the total pressure inside a tennis ball reaches 1013 kPa, the tennis ball will burst. If the ball
originally has a pressure of 405 kPa at room temperature (20°C), at what temperature will the ball burst?
(submit your work)

Respuesta :

Answer:

50.0 °C

Explanation:

If all other variables are being held constant, you can find the temperature using Guy-Lussac's Law. The equation looks like this:

P₁ / T₁ = P₂ / T₂

In this equation, "P₁" and "T₁" represent the initial pressure and temperature. "P₂" and "T₂" represent the final pressure and temperature. You can find the missing temperature (T₂) by plugging the given values into the equation and simplifying.

P₁ = 405 pKa                    P₂ = 1013 pKa

T₁ = 20 °C                         T₂ = ? °C

P₁ / T₁ = P₂ / T₂                                              <----- Guy-Lussac's Law

405 pKa / 20 °C = 1013 pKa / T₂                  <----- Insert values

20.25 = 1013 pKa / T₂                                   <----- Divide 20 from 405

20.25 x T₂ = 1013 pKa                                  <----- Multiply both sides by T₂

T₂ = 50.0 °C                                                  <----- Divide both sides by 20.25