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mestny [16]
3 years ago
5

A sample consisting of 2 moles He is expanded isothermally at 0 degrees from 5.0dm3 to 20.0dm3. Calculate w, q and deltaU for ea

ch of the following situations: (i) A reversible expansion of the sample. (ii) An irreversible expansion of the sample against a constant external pressure equal to the final pressure of the gas. (iii) A free expansion (against zero external pressure i.e. in a vacuum) of the sample.
Chemistry
1 answer:
FinnZ [79.3K]3 years ago
7 0

Answer:

i) \Delta U=0 w=-6293 J q=6293 J

ii) \Delta U=0 w=-3404,52 J q=3404,52 J

ii) \Delta U=0 w=0 J q=0 J

Explanation:

As the initial and final states of the sample are the same, the ΔU of the sample is, for the three cases

\Delta U=n.C_{V}.\Delta T=0 since \Delta T=0

i)Reversibly P_{ext} =P_{sys} so w can be calculated by  

w=-n.R.T.ln(\frac{V_{f}}{V_{i}})=-2 \times 8.314\frac{J}{mol K} \times 273,15K \times ln(\frac{}{5dm^{3}})=-6293 J

and because of the first law of thermodynamics

q=-w=6293 J

ii)Irreversibly with P_{ext} =P_{f}

we can calculate P_{f} by the law of ideal gases

P_{f} =\frac{n\times R\times T}{V_{f}} =\frac{2\times 0.082\frac{dm^{3}atm}{mol K}\times 273,15K}{20dm^{3}} =2,24 atm

then w can be calculated by

w=-P_{ext} \times \Delta V=-2,24 atm \times (20-5) dm^{3} \times frac{101.325J }{atm dm^{3}=-3404,52J

and  

q=-w=3404,52J

iii)a free expansion P_{ext} so w=0 (there's no work at vaccum) and q=-w=0

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8 0
3 years ago
A 2.5% (by mass) solution concentration signifies that there is of solute in every 100 g of solution. 2. therefore, when 2.5% is
densk [106]
Answer #1 is "there is 2.5 grams of solute in every 100 g of solution." 
We calculate for 2.5% by mass solution by dividing the mass of the solute by the mass of the solution and then multiply by 100.
Answer #2 is "that mass ratio would be 2.5/100 or 2.5 grams of solute/100 grams of solution." 
We weigh out 2.5 grams of solute and then add 97.5 grams of solvent to make a total of 100 gram solution, that is,
     mass of solute / mass of solution = 2.5g solute / (2.5g solute + 97.5g solvent)
                                                          = 2.5g solute / 100g solution
Answer#3 is "a solution mass of 1 kg is 10 times greater than 100 g, thus one kilogram (1 kg) of a 2.5% ki solution would contain 25 grams of ki."
We multiply 10 to each mass so that 100 grams becomes 1000grams since 1000 grams is equal to 1 kg:
     mass of solute / mass of solution = 2.5g*10/[(2.5g*10) + (97.5g*10)]
                                                          = 25g solute/(25g solute + 975g solvent)
                                                          = 25g solute/1000g solution
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5 0
3 years ago
Read 2 more answers
3. Determine the moles of sodium, Na, containing 7.9x1024 atoms.​
-BARSIC- [3]

Answer:

12.7mol Na.

Explanation:

Hello there!

In this case, according to the concept of mole, which stands for the amount of substance, we can recall the concept of Avogadro's number whereby we understand that one mole of any substance contains 6.022x10²³ particles, for the given atoms of sodium, we can calculate the moles as shown below:

7.9x10^{23}atoms*\frac{1mol}{6.022x10^{23}atoms} \\\\

Thus, by performing the division we obtain:

12.7molNa

Regards!

8 0
3 years ago
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