Science & Mathematics - 12 December 2006

2 Cases:

(-1000J energy lost in each case)= q

1 - Piston can move, change in volume occurs
...... w= P*deltaV... Work is positive. (WHY?)
2 - Piston can not move, no change in volume.
...... w=P*0 = 0

What can you say about U?

U = q + w

(1) - Change in U is less negative. (WHY?)
(2) - Since w = 0, Change in U = q = -1000J

______

In which case does the gas have a higher temperature after the removal of the energy?

Answer: deltaT and deltaU are related (HOW?) ... if deltaU is smaller, deltaT is smaller. ... So case 1 has lower temperature. (WHY?)


These are the parts of the question i dont get.


Note: The answers I gave are correct... The Why and Hows aren't asked of us on the test, they are just me personally not understanding why those answers are correct.

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Thanks everyone

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