Question

In a sealed and rigid container, a sample of gas at 2.50 atm and 60.0 °C is cooled to 20.0 °C. What is the pressure (in atm) of the gas at 20.0 °C?

Answer 1

Using Gay-Lussac's law, the pressure of the gas at 20.0 °C is calculated to be 2.20 atm after converting Celsius to Kelvin and applying the formula P1/T1 = P2/T2.

Step-by-step explanation:

To find the pressure of the gas after cooling, we can apply Gay-Lussac's law which relates the pressure of a gas to its temperature, as long as the volume and amount of gas remain constant. The law is usually stated as P1/T1 = P2/T2, where P1 and P2 are the initial and final pressures, and T1 and T2 are the initial and final absolute temperatures (in kelvins), respectively.

First, convert Celsius temperatures to Kelvin: T1 = 60.0 °C + 273.15 = 333.15 K and T2 = 20.0 °C + 273.15 = 293.15 K. Then we plug the values into Gay-Lussac’s law: (2.50 atm / 333.15 K) = (P2 / 293.15 K). Solving for P2 gives us the final pressure of the gas at 20.0 °C.

By cross-multiplying and dividing, we find P2 = 2.50 atm * (293.15 K / 333.15 K) = 2.20 atm. Therefore, the pressure of the gas at 20.0 °C is 2.20 atm.