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?
Chemistry (please help!!) - Celine, Thursday, March 24, 2011 at 12:20am
My answer is 594 K... which is excessively high...
Chemistry (please help!!) - Celine, Thursday, March 24, 2011 at 12:23am
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...
Chemistry (please help!!) - Liz, Thursday, March 24, 2011 at 8:39am
(P1xV1)/T1 = (P2xV2)/T2
T2 = (P2xV2xT1)/(P1xV1)
Chemistry (please help!!) - Celine, Thursday, March 24, 2011 at 12:08pm
That's the equation I use and I get 594 K... I think that's a really high value, is that answer right?
To solve this problem, you can use the ideal gas law equation: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature.
In this case, we have two different situations: before and after the bag is inflated. We can use the ideal gas law equation to solve for the temperature in each case.
Before the bag is inflated, the pressure is 5.7 kPa and the volume is 0.75 L. The number of moles and the gas constant are not given, but we can assume they are constant for both situations.
So we have: (P1 x V1) / T1 = (P2 x V2) / T2
Substituting the values given: (5.7 kPa x 0.75 L) / (24°C in Kelvin) = (7.2 kPa x 1.2 L) / T2
To convert the temperature to Kelvin, you add 273.15 to the Celsius temperature. So 24°C = 24 + 273.15 = 297.15 K
Simplifying the equation, we get: (4.275 kPa L / 297.15 K) = (8.64 kPa L) / T2
Now solve for T2:
T2 = (8.64 kPa L) / (4.275 kPa L / 297.15 K)
T2 = (8.64 kPa L) x (297.15 K / 4.275 kPa L)
T2 = 594 K (approximately, after rounding)
So the temperature outside when the paper bag explodes is approximately 594 Kelvin.
It seems like your initial answer of 594 K is correct. However, it is always a good idea to double-check your calculations to ensure accuracy.
No, the answer of 594 K is not correct. It seems there might be a mistake in your calculations. Let's go through the steps again to find the correct solution.
We can use the ideal gas law equation to solve this problem:
(P1 x V1) / T1 = (P2 x V2) / T2
Given:
P1 = 5.7 kPa
V1 = 0.75 L
T1 = 24°C
P2 = 7.2 kPa
V2 = 1.2 L
T2 = ?
First, let's convert the temperature from Celsius to Kelvin:
T1 = 24°C + 273.15 = 297.15 K
Now we can plug in the values into the equation:
(5.7 kPa x 0.75 L) / 297.15 K = (7.2 kPa x 1.2 L) / T2
Simplifying:
4.275 kPa L / 297.15 K = 8.64 kPa L / T2
Cross multiply:
(4.275 kPa L) x T2 = (8.64 kPa L) x 297.15 K
Divide both sides by 4.275 kPa L:
T2 = (8.64 kPa L x 297.15 K) / 4.275 kPa L
T2 ≈ 60.2 K
So, the temperature outside is approximately 60.2 Kelvin (K).