A stressed chemistry student is hyperventilating into a paper bag. The student stops to wipe his brow after a few breaths, and the bag is at 5.7 KPa and the temperature in the lecture hall is 24 °C. The bag is 0.75L full. The student then goes at it again, and completely fills the bag with air to 1.2L and 7.2KPa. The student panics, and runs out of the lecture hall...when he gets outside, the inflated paper bag explodes. What is the temperature outside?
What exactly does 0.75L full mean? The bag is 3/4 full or what?
No I don't think so because after the student goes at it again it is filled to 1.2L and it can't be more than completely filled. I think I am probably supposed to use V1/T1=V2/T2 or P1V1/T1= P2V2/T2. I just do not know because I feel like I have two volumes.
To solve this problem, we can use the combined gas law, which incorporates the relationships between pressure, volume, and temperature.
The combined gas law equation is:
(P1 * V1) / T1 = (P2 * V2) / T2
Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure
V2 = final volume
T2 = final temperature
Let's list the given values:
Initial conditions (inside the lecture hall):
P1 = 5.7 KPa
V1 = 0.75 L
T1 = 24 °C
Final conditions (outside the lecture hall):
P2 = 7.2 KPa
V2 = 1.2 L
T2 = ? (what we need to find)
Now we can plug these values into the combined gas law equation and solve for T2.
(5.7 * 0.75) / (24 + 273) = (7.2 * 1.2) / (T2 + 273)
Let's simplify the equation:
4.275 / (24 + 273) = 8.64 / (T2 + 273)
Cross-multiplying:
4.275 * (T2 + 273) = 8.64 * (24 + 273)
4.275T2 + 4.275 * 273 = 8.64 * 297
Now, solve for T2:
4.275T2 = (8.64 * 297) - (4.275 * 273)
T2 = [(8.64 * 297) - (4.275 * 273)] / 4.275
T2 ≈ 26.5 °C
Therefore, the temperature outside, where the inflated paper bag exploded, is approximately 26.5 °C.