Physical Chemistry- Partial Pressures

AI Thread Summary
The discussion revolves around a homework problem involving the decomposition of KCLO3 and the collection of oxygen gas over water. The student initially miscalculated the number of moles of oxygen, leading to an incorrect final volume. After receiving feedback, they corrected the arithmetic error and recalculated the number of moles and the volume of oxygen at standard conditions. The final volume was confirmed to be approximately 25.98 cm³, indicating that the student ultimately approached the problem correctly after addressing the mistake. The importance of double-checking calculations for accuracy was emphasized throughout the discussion.
PerenialQuest
Messages
9
Reaction score
0
Hello all,
This is a homework problem for my CHE345 class. Not sure what to do here, please at least let me know if I'm in the right ballpark.

Homework Statement



A student decomposes KCLO3 and collects 35.2 cm^3 of O2 over water at 23.0°C. The laboratory barometer reads 751 Torr. The vapor pressure of water at 23.0°C is 21.1 Torr. Find the volume the oxygen would occupy at 0.0*C and 1.2atm (= 912 Torr).

Homework Equations


not exactly sure what is relevant here. Perhaps these:
(1) PV = nRT
(2) Pj = partial pressure
(3) xj = mole fraction
(4) Pj = xjP
(5) Pj = njRT/V
(6) Ptotal = PA + PB +...

The Attempt at a Solution


My problem I think, is in interpreting the question, but here goes an attempt:

using eq. (6):
Ptotal = Pwater + PO2
751 Torr = 21.1 Torr + PO2
PO2 = 729.9

from here I'll use eq. (1) to solve for nO2
PV = nRT
n = PV/RT
n = (729.9 Torr)(0.0352 L)/(62.634 L Torr K-1mol-1)(296.15 K)
n = 1.385 x 10-4

Now to solve for the V at Tfinal

V = nRT/P
V = (1.385E-4 moles)(62.634 L Torr K-1mol-1)(273.15 K) / 912 Torr)
V = 0.002598 L = 2.598cm3

How'd I do?
 
Physics news on Phys.org
PerenialQuest said:
Hello all,
This is a homework problem for my CHE345 class. Not sure what to do here, please at least let me know if I'm in the right ballpark.

Homework Statement



A student decomposes KCLO3 and collects 35.2 cm^3 of O2 over water at 23.0°C. The laboratory barometer reads 751 Torr. The vapor pressure of water at 23.0°C is 21.1 Torr. Find the volume the oxygen would occupy at 0.0*C and 1.2atm (= 912 Torr).

Homework Equations


not exactly sure what is relevant here. Perhaps these:
(1) PV = nRT
(2) Pj = partial pressure
(3) xj = mole fraction
(4) Pj = xjP
(5) Pj = njRT/V
(6) Ptotal = PA + PB +...

The Attempt at a Solution


My problem I think, is in interpreting the question, but here goes an attempt:

using eq. (6):
Ptotal = Pwater + PO2
751 Torr = 21.1 Torr + PO2
PO2 = 729.9

from here I'll use eq. (1) to solve for nO2
PV = nRT
n = PV/RT
n = (729.9 Torr)(0.0352 L)/(62.634 L Torr K-1mol-1)(296.15 K)
n = 1.385 x 10-4

Now to solve for the V at Tfinal

V = nRT/P
V = (1.385E-4 moles)(62.634 L Torr K-1mol-1)(273.15 K) / 912 Torr)
V = 0.002598 L = 2.598cm3

How'd I do?

You made an arithmetic error in calculating the number of moles. The pressures and temperatures in the two situations are not too different, and, if the volume was much less than 35.2 cc, this should have alerted you that there was an arithmetic mistake somewhere. Try to get used to asking yourself whether your answer makes sense.
 
Chestermiller said:
You made an arithmetic error in calculating the number of moles. The pressures and temperatures in the two situations are not too different, and, if the volume was much less than 35.2 cc, this should have alerted you that there was an arithmetic mistake somewhere. Try to get used to asking yourself whether your answer makes sense.

1. Homework Statement

A student decomposes KCLO3 and collects 35.2 cm^3 of O2 over water at 23.0°C. The laboratory barometer reads 751 Torr. The vapor pressure of water at 23.0°C is 21.1 Torr. Find the volume the oxygen would occupy at 0.0*C and 1.2atm (= 912 Torr).

2. Homework Equations
not exactly sure what is relevant here. Perhaps these:
(1) PV = nRT
(2) Pj = partial pressure
(3) xj = mole fraction
(4) Pj = xjP
(5) Pj = njRT/V
(6) Ptotal = PA + PB +...

3. The Attempt at a Solution
My problem I think, is in interpreting the question, but here goes an attempt:

using eq. (6):
Ptotal = Pwater + PO2
751 Torr = 21.1 Torr + PO2
PO2 = 729.9

from here I'll use eq. (1) to solve for nO2
PV = nRT
n = PV/RT
n = (729.9 Torr)(0.0352 L)/(62.634 L Torr K-1mol-1)(296.15 K)
n = 1.385 x 10-3

Now to solve for the V at Tfinal

V = nRT/P
V = (1.385E-3 moles)(62.634 L Torr K-1mol-1)(273.15 K) / 912 Torr)
V = 0.02598 L = 25.98cm3

Ah yes, thank you, I missed a digit. That better? Did I go about the problem correctly?
 
Yes. Nice job.
 
I don't get how to argue it. i can prove: evolution is the ability to adapt, whether it's progression or regression from some point of view, so if evolution is not constant then animal generations couldn`t stay alive for a big amount of time because when climate is changing this generations die. but they dont. so evolution is constant. but its not an argument, right? how to fing arguments when i only prove it.. analytically, i guess it called that (this is indirectly related to biology, im...
Back
Top