Question
A 3.5 liter sample at 20 ºC and a pressure of 650 mmHg is allowed to expand to a volume of 8570 mL. The final pressure of the gas is 0.75 atm. What is the final temperature in ºC?
Answers
GPT 3.5
To solve this problem, we can use the combined gas law, which states that for a given amount of gas at constant mass, the ratio of the pressure and volume is directly proportional to the ratio of the temperature:
(P1)(V1) / T1 = (P2)(V2) / T2
where:
P1 = initial pressure = 650 mmHg
V1 = initial volume = 3.5 L = 3500 mL
T1 = initial temperature = 20 ºC + 273.15 = 293.15 K
P2 = final pressure = 0.75 atm
V2 = final volume = 8570 mL
T2 = final temperature = ? ºC
Plugging in the values:
(650 mmHg)(3500 mL) / 293.15 K = (0.75 atm)(8570 mL) / T2
Solving for T2:
(650)(3500) = (0.75)(8570)(293.15)
T2 = (0.75)(8570)(293.15) / (650)(3500)
T2 = 295.59 K
Converting back to ºC:
T2 = 295.59 K - 273.15 = 22.44 ºC
Therefore, the final temperature of the gas is approximately 22.44 ºC.
(P1)(V1) / T1 = (P2)(V2) / T2
where:
P1 = initial pressure = 650 mmHg
V1 = initial volume = 3.5 L = 3500 mL
T1 = initial temperature = 20 ºC + 273.15 = 293.15 K
P2 = final pressure = 0.75 atm
V2 = final volume = 8570 mL
T2 = final temperature = ? ºC
Plugging in the values:
(650 mmHg)(3500 mL) / 293.15 K = (0.75 atm)(8570 mL) / T2
Solving for T2:
(650)(3500) = (0.75)(8570)(293.15)
T2 = (0.75)(8570)(293.15) / (650)(3500)
T2 = 295.59 K
Converting back to ºC:
T2 = 295.59 K - 273.15 = 22.44 ºC
Therefore, the final temperature of the gas is approximately 22.44 ºC.