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A cylinder with a movable piston contains 0.1 mole of a monoatomic ideal gas. The piston moves through state a, b and c. The heat Q, changes from state c to a is 685 J. The work W, changes from state c to a is -120 J. The work, W performed from state a to b then to c is 75 J. By using the first law of thermodynamic, U = Q W where U is the internal energy : (a) Determine the change in internal energy between states a and c. (b) Is heat added or removed from the gas when the gas is taken along the path abc? (c) Calculate the heat added or removed when the gas is taken along the path abc?
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Answer Posted / kang chuen tat (malaysia - pen
(a) U = Q + W = 685 J – 120 J = 565 J. (b) For path a to c, U = -565 J – heat is removed. (c) Q = U – W = -565 J – 75 J = -640 J (heat energy is removed when the value is negative). The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.
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