HW #7.45, pg. 282

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Shannon Han 2B
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Joined: Fri Sep 26, 2014 2:02 pm

HW #7.45, pg. 282

Postby Shannon Han 2B » Mon Jan 12, 2015 4:40 pm

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?

Kayla Denton 1A
Posts: 106
Joined: Fri Sep 26, 2014 2:02 pm

Re: HW #7.45, pg. 282

Postby Kayla Denton 1A » Mon Jan 12, 2015 5:05 pm

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.

Kayla Denton 1A
Posts: 106
Joined: Fri Sep 26, 2014 2:02 pm

Re: HW #7.45, pg. 282

Postby Kayla Denton 1A » Mon Jan 12, 2015 5:09 pm

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.

Ishkhan 3O
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Joined: Fri Sep 26, 2014 2:02 pm

Re: HW #7.45, pg. 282

Postby Ishkhan 3O » Sun Jan 18, 2015 6:48 pm

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.

Ishkhan 3O
Posts: 24
Joined: Fri Sep 26, 2014 2:02 pm

Re: HW #7.45, pg. 282

Postby Ishkhan 3O » Sun Jan 18, 2015 7:50 pm

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?

Neil DSilva 1L
Posts: 70
Joined: Fri Sep 26, 2014 2:02 pm

Re: HW #7.45, pg. 282

Postby Neil DSilva 1L » Sun Jan 18, 2015 8:33 pm

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.


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