a stressed chemistry student is hyperventilating into a paper bag. The student stops to wip his brow after a few breaths, and the bag is at 5.7 KPa and the temperature in the lecture hall is 24 degrees C. The bag is 0.75 L full. The student then goes at it again, and completely fills the bag with air to 1.2 L and 7.2KPa. The student panics, and runs out of the lecture hall... when he gest outside, the inflated paper bag explodes. What is the temperature outside?
My answer is 594 K... which is excessively high...
Do I not need to turn the temperature to Kelvin for the equation?? If I don't turn it to K I get 49 C... which is a more reasonable value... I left the other values with their same units... please someone answer...
(P1xV1)/T1 = (P2xV2)/T2
T2 = (P2xV2xT1)/(P1xV1)
That's the equation I use use and I get 594 K... I think that's a really high value, is that answer right?
To determine the temperature outside when the inflated paper bag explodes, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.
First, let's find the initial number of moles of gas in the bag when it is at 5.7 KPa and 0.75 L using the ideal gas law. We'll assume the bag contains only air, which is mostly nitrogen and oxygen.
Step 1: Convert the initial pressure to atmospheres (1 atm = 101.325 KPa).
5.7 KPa / 101.325 KPa/atm ≈ 0.0563 atm
Step 2: Convert the volume to liters.
0.75 L
Step 3: Determine the number of moles of gas using the ideal gas law.
n = PV / RT
n = (0.0563 atm) * (0.75 L) / (0.0821 L·atm/(mol·K) * (24 + 273.15) K
Step 4: Calculate the number of moles.
n ≈ 0.0017 moles
Now, let's find the final temperature outside when the bag is inflated to 1.2 L and 7.2 KPa.
Step 1: Convert the final pressure to atmospheres.
7.2 KPa / 101.325 KPa/atm ≈ 0.071 atm
Step 2: Convert the volume to liters.
1.2 L
Step 3: Determine the number of moles using the ideal gas law.
n = PV / RT
n = (0.071 atm) * (1.2 L) / (0.0821 L·atm/(mol·K) * T
Step 4: Substitute the initial moles obtained earlier.
0.0017 moles = (0.071 atm) * (1.2 L) / (0.0821 L·atm/(mol·K) * T
Step 5: Solve for T (temperature).
T ≈ (0.071 atm) * (1.2 L) / (0.0821 L·atm/(mol·K) * 0.0017 moles
Calculating this expression results in T ≈ 1928 K (Kelvin).
Finally, to convert this temperature from Kelvin to Celsius, subtract 273.15 from 1928.
1928 K - 273.15 ≈ 1654.85 degrees Celsius.
Therefore, the temperature outside when the inflated paper bag explodes is approximately 1654.85 degrees Celsius.