High School

A sealed container holding 0.0255 L of an ideal gas at 0.983 atm and 65.1 °C is placed into a refrigerator and cooled to 39.3 °C with no change in volume. Calculate the final pressure of the gas.

Answer :

After converting the initial and final temperatures to Kelvin and applying the combined gas law (P1/T1 = P2/T2), it is determined that the final pressure of the gas after cooling to 39.3 °C at constant volume is 0.905 atm.

To calculate the final pressure of an ideal gas when it is cooled from 65.1 °C to 39.3 °C at constant volume, we can use the combined gas law which relates pressure, volume, and temperature:

P1/T1 = P2/T2

Here, P1 and T1 are the initial pressure and temperature, and P2 and T2 are the final pressure and temperature.

First, we need to convert the temperatures from degrees Celsius to Kelvin:

  • T1 = 65.1
    °C + 273.15 = 338.25 K
  • T2 = 39.3
    °C + 273.15 = 312.45 K

Now, using the initial conditions:

  • P1 = 0.983 atm
  • T1 = 338.25 K

And the given final temperature (T2), we can solve for the final pressure (P2):

P2 = P1 * (T2/T1)

P2 = 0.983 atm * (312.45 K / 338.25 K)

Calculate the final pressure:

P2 = 0.905 atm (rounded to three decimal places)

Thus, the final pressure of the gas after being cooled to 39.3
°C is 0.905 atm.

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