The Relationship Between Pressure, Volume, and Temperature in Ideal Gas Law
Question:
I have a cylinder-shaped glass container that is 15 cm tall. It holds 100 cm^3 of nitrogen gas at 1 atm pressure (101.3 kPa) and 22 degrees C. I place a rubber stopper in the top so that no gas can escape. If a 40 N force is required to force the rubber stopper off the top of the flask, what temperature can I heat the nitrogen to with a Bunsen burner before the rubber stopper pops off? (The answer should be in units of Kelvin)
Answer:
The nitrogen gas can be heated to 295 K (22 degrees C) before the rubber stopper pops off.
To determine the temperature at which the rubber stopper will pop off, we need to consider the relationship between pressure, volume, and temperature (PVT) of the nitrogen gas.
The ideal gas law states that: PV = nRT
where:
- P is the pressure in kPa
- V is the volume in cm^3
- n is the number of moles of gas
- R is the ideal gas constant (8.314 J/mol·K)
- T is the temperature in Kelvin
We are given the following initial conditions:
- P = 101.3 kPa
- V = 100 cm^3
- T = 295 K (22 degrees C)
Next, we can use Boyle's law to determine the pressure at the point where the stopper pops off:
P1V1 = P2V2
where:
- P1 = 101.3 kPa (initial pressure)
- V1 = 100 cm^3 (initial volume)
- V2 = 100 cm^3 (volume remains constant)
Solving for P2, we get:
P2 = (P1V1) / V2 = (101.3 kPa * 100 cm^3) / 100 cm^3 = 101.3 kPa
Finally, we can use Charles's law to determine the temperature at which the pressure reaches 101.3 kPa:
V1 / T1 = V2 / T2
where:
- T1 = 295 K (initial temperature)
- T2 is the temperature at which the pressure reaches 101.3 kPa
Solving for T2, we get:
T2 = (T1 * V1) / V2 = (295 K * 100 cm^3) / 100 cm^3 = 295 K
Therefore, the nitrogen gas can be heated to 295 K (22 degrees C) before the rubber stopper pops off.