Self Test 7.6A How do we Find the Original Volume?


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Jennifer Aguayo 1I
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Self Test 7.6A How do we Find the Original Volume?

Postby Jennifer Aguayo 1I » Sun Jan 18, 2015 10:38 pm

Suppose that 2.00mol CO2 at 2.00atm and 300.K is compressed isothermally and reversibly to half its original volume before being used to produce soda water. Calculate w, q, and delta U by treating the CO2 as an ideal gas.

I am confused on how to find the original volume to solve for w. Or do we not need it?

Thank you!

Kayla Denton 1A
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Re: Self Test 7.6A How do we Find the Original Volume?

Postby Kayla Denton 1A » Sun Jan 18, 2015 10:59 pm

You don't need it! :)
If you use the equation w = -nRTln(Vf/Vi), all you need to know is that the final volume is half of the original volume, so the ratio Vf/Vi is 1/2. You have n, R, and T, so you can calculate w.
Hope this answers your question!

The rest of the question is below just in case.
w = -(2.00 mol)(8.3145 J/Kmol)(300 K)(1/2) = +3460 J = +3.46 kJ.
Then you can calculate delta U and q.
Change in internal energy is 0 for isothermal expansions.
Delta U = q + w
0 = q + w
q = -w
q = -3.46 kJ.

Neil DSilva 1L
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Re: Self Test 7.6A How do we Find the Original Volume?

Postby Neil DSilva 1L » Sun Jan 18, 2015 11:07 pm

Even though you don't need to find the exact initial volume for this problem, it's helpful to remember that since you're dealing with an ideal gas you can just use PV=nRT and the given initial conditions to find any unknown conditions. In this case, you're given P, T, and n (and you always have R). Plug these values into V=nRT/P and you will have the initial volume.

Justin Le 2I
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Re: Self Test 7.6A How do we Find the Original Volume?

Postby Justin Le 2I » Sat Jan 24, 2015 8:41 pm

Adding on to the previous comment, if you notice that you know three of the values, then there is a good chance that you will have to use PV=nRT.

Chem_Mod
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Re: Self Test 7.6A How do we Find the Original Volume?

Postby Chem_Mod » Sat Jan 31, 2015 1:50 am

Kayla is correct. The most advantageous part of any of the formulas that have logs is that you can specify a ratio, such as "half the volume" or "three times the pressure" instead of actual values.


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