Question
A sample of oxygen gas exerts a pressure of 500.0 kPa against the walls of its rigid container at 20.0 °C. At what temperature will the pressure rise to 1,000.0 kPa?
• 313 °C
10 °c
• 40 °c
586 °C
• 313 °C
10 °c
• 40 °c
586 °C
Answers
GPT 3.5
To solve this problem, we can use the ideal gas law:
PV = nRT
Where:
P = pressure
V = volume
n = moles of gas
R = ideal gas constant
T = temperature in Kelvin
Since the volume and moles of gas are constant in this case, we can simplify the equation to:
P1/T1 = P2/T2
Where:
P1 = initial pressure (500.0 kPa)
T1 = initial temperature (20.0 °C + 273 = 293 K)
P2 = final pressure (1000.0 kPa)
T2 = final temperature (unknown)
Now we can plug in the values and solve for T2:
(500.0 kPa) / (293 K) = (1000.0 kPa) / T2
T2 = (1000.0 kPa * 293 K) / 500.0 kPa
T2 = 586 K
Converting this temperature to Celsius:
T2 = 586 K - 273 = 313°C
Therefore, the temperature at which the pressure will rise to 1000.0 kPa is 313 °C.
PV = nRT
Where:
P = pressure
V = volume
n = moles of gas
R = ideal gas constant
T = temperature in Kelvin
Since the volume and moles of gas are constant in this case, we can simplify the equation to:
P1/T1 = P2/T2
Where:
P1 = initial pressure (500.0 kPa)
T1 = initial temperature (20.0 °C + 273 = 293 K)
P2 = final pressure (1000.0 kPa)
T2 = final temperature (unknown)
Now we can plug in the values and solve for T2:
(500.0 kPa) / (293 K) = (1000.0 kPa) / T2
T2 = (1000.0 kPa * 293 K) / 500.0 kPa
T2 = 586 K
Converting this temperature to Celsius:
T2 = 586 K - 273 = 313°C
Therefore, the temperature at which the pressure will rise to 1000.0 kPa is 313 °C.
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