Question

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.

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

Answer 2

Answer:

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