Question

A rigid ball that contained 25.0 liters of air at 22 degrees c and 6.25 atm pressure was placed in an oven at a temperature of 100 degrees c. What was the new pressure inside the ball?

Answer 1

Using Gay-Lussac's Law, the new pressure inside a ball heated to 100 degrees Celsius is approximately 7.9 atm.

Step-by-step explanation:

The question you've asked involves the behavior of gas under different conditions of temperature and pressure, which can be described using the Gay-Lussac's Law of combining volumes from the ideal gas law. This law states that the pressure of a given mass of gas is directly proportional to its absolute temperature (in Kelvin) when the volume is kept constant.

To find the new pressure inside the ball when it's placed in the oven, we use the following form of Gay-Lussac's Law:

P1/T1 = P2/T2

Where,

P1 is the initial pressure (6.25 atm)T1 is the initial temperature in Kelvin (22 degrees Celsius + 273.15 = 295.15 K)P2 is the final pressure (which we are looking for)T2 is the final temperature in Kelvin (100 degrees Celsius + 273.15 = 373.15 K)

By rearranging the formula to solve for P2, we get:

P2 = P1 × (T2/T1)

P2 = 6.25 atm × (373.15 K / 295.15 K)

Now let's plug the values into the equation:

P2 = 6.25 atm × 1.264 = 7.9 atm (approximately)

The new pressure inside the ball when placed in an oven at 100 degrees Celsius is approximately 7.9 atm.