Physical Chemistry- Partial Pressures

[PerenialQuest]

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) P_j = partial pressure
(3) x_j = mole fraction
(4) P_j = x_jP
(5) P_j = n_jRT/V
(6) P_total = P_A + P_B +...

The Attempt at a Solution

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

using eq. (6):
P_total = P_water + P_O2
751 Torr = 21.1 Torr + P_O2
P_O2 = 729.9

from here I'll use eq. (1) to solve for n_O2
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 T_final

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?

 

[Chestermiller]

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) P_j = partial pressure
(3) x_j = mole fraction
(4) P_j = x_jP
(5) P_j = n_jRT/V
(6) P_total = P_A + P_B +...

The Attempt at a Solution

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

using eq. (6):
P_total = P_water + P_O2
751 Torr = 21.1 Torr + P_O2
P_O2 = 729.9

from here I'll use eq. (1) to solve for n_O2
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 T_final

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.

 

[PerenialQuest]

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?

 

[Chestermiller]

Yes. Nice job.