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ss7ja [257]
3 years ago
13

Calculate the final temperature of a sample of carbon dioxide of mass 16.0 g that is expanded reversibly and adiabatically from

500 cm3 at 298.15 K to 2.00 dm3. 1) If you want to calculate the work, dW=-pdV -> W = -nRTln(V2/V1). Is this expression ok for this problem? If yes, then what T should I use? 1) Why the Charles law V1/T1=V2/T2 does not hold? If you increase V you do work and so you expect to reduce U and therefore T. But according to charles law, in order to keep the proportion, if you increase V, you also increase T.
Chemistry
1 answer:
Pie3 years ago
4 0

For your first question, that equation only works if your situation is occurring at a constant temperature. Your original question is such a situation - everything occurs at 298.15 K. Therefore, you can use this value in the equation to calculate work. For your second question, Charles' Law describes how the volume of gas changes as you heat or cool it, PROVIDED PRESSURE AND MOLES OF GAS REMAIN CONSTANT THE WHOLE TIME. In your original question above, temperature stays constant while volume changes. However, what they don't tell you is that this necessarily requires a change in either pressure or moles of gas. Because the question works with the same sample the of gas the whole time (i.e. moles are constant), it is pressure that is changing (and this change will occur according to Boyle's Law, since temperature and moles are held constant). Hope that clarifies things!

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A warm air is rising as cooler air es in to take its place. what is the name for that?​
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Explanation:

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Jen makes a Venn diagram to compare active transport and passive transport.
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Read 2 more answers
How many number of moles are present in 200. g of sodium bicarbonate (NaHCO3)
nikdorinn [45]

You can solve this problem through dimensional analysis.

First, find the molar mass of NaHCO3.

Na = 22.99 g

H = 1.008 g

C = 12.01 g

O (3) = 16 (3) g

Now, add them all together, you end with with the molar mass of NaHCO3.

22.99 + 1.008 + 12.01 + 16(3) = 84.008 g NaHCO3. This number means that for every mole of NaHCO3, there is 84.008 g NaHCO3. In simpler terms, 1 mole NaHCO3 = 84.008 g NaHCO3.

After finding the molar mass of sodium bicarbonate, now you can use dimensional analysis to solve for the number of moles present in 200. g of sodium bicarbonate.

200. g NaHCO_3 * \frac{1 mole NaHCO_3}{84.008 g NaHCO_3}

Cross out the repeating units which are g NaHCO3, and the remaining unit is mole NaHCO3

200.  * 1 = 200

200/ 84.008 = 2.38

Notice how there are only 3 sig figs in the answer. This is because the given problem only gave three sig figs.

Your final answer is 2.38 mol NaHCO3.

8 0
3 years ago
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