For #45 on the homework, the question asks to estimate the mass of octane that would need to be burned to product enough heat to raise the temperature of the air in a 12x12x8 ft room from 40 degrees F to 78 degrees F using the normal composition of air to determine its density and assuming a pressure of 1.00 atm. I'm confused on the calculations of d. Where did the solutions manual get 277.6 K for T? Then, when I used that value for T in the equation d = PM/RT, I got 1.27, not 1.27*10^-3.
Also, on part b of the question, how do we know that 1 gal = 3.785*10^3 mols of octane? Where did the 2 mols of octane per -10,942 kJ come from? I'm confused where all these numbers came from. Could anyone help explain the calculations for these two parts of 45?
HW #7.45, pg. 282
Moderators: Chem_Mod, Chem_Admin
-
- Posts: 106
- Joined: Fri Sep 26, 2014 2:02 pm
Re: HW #7.45, pg. 282
Hi! I posted a video explaining this question here:
viewtopic.php?f=76&t=4968
You can just scroll through to find the answers to your questions!
To quickly answer your first two questions: 277.6 K is the initial temperature, and you probably got 1.27 because you thought that 1 L = 1 m^3 but really 1 L = 1 dm^3.
viewtopic.php?f=76&t=4968
You can just scroll through to find the answers to your questions!
To quickly answer your first two questions: 277.6 K is the initial temperature, and you probably got 1.27 because you thought that 1 L = 1 m^3 but really 1 L = 1 dm^3.
-
- Posts: 106
- Joined: Fri Sep 26, 2014 2:02 pm
Re: HW #7.45, pg. 282
My video doesn't explain part b so I'll explain it here.
You need to look up the conversion between gal to mL. 1.0 gal = 3.785 x 10^3 mL.
Also, note that the molar mass of octane is 114.232 g/mol.
Next:
1.0 gal x (3.785 x 10^3 mL/gal) x (0.70 g octane / mL) x (mol octane/114.232 g) x (-5471 kJ/mol [from reaction]) = -1.3 x 10^5 kJ of heat produced. The units all cancel out so you're left with kJ.
You need to look up the conversion between gal to mL. 1.0 gal = 3.785 x 10^3 mL.
Also, note that the molar mass of octane is 114.232 g/mol.
Next:
1.0 gal x (3.785 x 10^3 mL/gal) x (0.70 g octane / mL) x (mol octane/114.232 g) x (-5471 kJ/mol [from reaction]) = -1.3 x 10^5 kJ of heat produced. The units all cancel out so you're left with kJ.
-
- Posts: 24
- Joined: Fri Sep 26, 2014 2:02 pm
Re: HW #7.45, pg. 282
For part A, you get the answer in g/L so you need to convert that to g/cm^3. The conversion is 1000 cm^3 to 1 L.
-
- Posts: 24
- Joined: Fri Sep 26, 2014 2:02 pm
Re: HW #7.45, pg. 282
In the solutions manual, when it calculates the heat required it has grams and moles in the same equation, but they don't cross out. Is that an error?
-
- Posts: 70
- Joined: Fri Sep 26, 2014 2:02 pm
Re: HW #7.45, pg. 282
Yes, the specific heat capacity for air should be 1.01 J/C.g (then grams cancels out). This is mentioned in the Solution Manual Errors file linked here: https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14B/Solution_Manual_Errors_5Ed.pdf. There are also a few other errors in the solutions manual for that problem.
Return to “Heat Capacities, Calorimeters & Calorimetry Calculations”
Who is online
Users browsing this forum: No registered users and 1 guest