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.

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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the answer is c i took the test