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In a sealed and rigid container, a sample of gas at 3.25 atm and 210.0 °C is cooled to 0.0 °C. What is the pressure (in atm) of the gas at 0.0 °C?

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Answer:

The pressure of the gas at 0.0 °C is 1.84 atm.

Step-by-step explanation:

To solve this problem, we can use the combined gas law, which states that the equation relates the pressure, volume, and temperature of a gas:

(P1 x V1) / T1 = (P2 x V2) / T2

where P1, V1, and T1 are the initial pressure, volume, and temperature, and P2, V2, and T2 are the final pressure, volume, and temperature.

We are given:

P1 = 3.25 atm

T1 = 210.0 °C = 483.15 K (convert to Kelvin)

T2 = 0.0 °C = 273.15 K (convert to Kelvin)

Since the container is rigid, the volume (V1) and final volume (V2) are the same. Therefore, we can simplify the equation to:

P1 / T1 = P2 / T2

Substituting the given values, we get:

(3.25 atm) / (483.15 K) = P2 / (273.15 K)

Simplifying, we get:

P2 = (3.25 atm x 273.15 K) / 483.15 K

P2 = 1.84 atm

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