A sealed container holding 0.0255 L of an ideal gas at 0.991 atm and 69 ∘ C is placed into a refrigerator and cooled to 43 ∘ C with no change in volume. Calculate the final pressure of the gas.

Question

Claire Arroyo

Chemistry

18 November, 13:02

A sealed container holding 0.0255 L of an ideal gas at 0.991 atm and 69 ∘ C is placed into a refrigerator and cooled to 43 ∘ C with no change in volume. Calculate the final pressure of the gas.

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Answers (2)

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Iris Cunningham · Answer 1

Answer: Final pressure of the gas is 0.916atm

Explanation: Using Charles law formula. P1V1T2 = P2V2T1

Where P1 = 0.991atm, V1 = 0.0255L, T1 = 69°C = 273+69 = 546.15, P2 = x, V2 = 0.0255L, T2 = 43°C = 273+43 = 316.15K.

Solution.

Substituting all the parameters into the equation

P1V1T2=P2V2T1,

we get,

0.991*0.0255*316.15 = P2 * 0.0255 * 546.15

P2 = 0.916atm

Juliette Sharp · Answer 2

The final pressure of the gas is 0.915atm

Explanation:

We have to apply the Charles Gay Lussac Law, where the pressure changes directly proportional to absolute T°

- No change in volume

- The same moles in both situations

P1 / T1 = P2 / T2

0.991 atm / 342K = P2 / 316k

(0.991 atm / 342K). 316K = P2

0.915 atm = P2