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