1.95 mol of an ideal gas at 300 K and 3.00 atm expands from 16 L to 28 L and a f
ID: 743012 • Letter: 1
Question
1.95 mol of an ideal gas at 300 K and 3.00 atm expands from 16 L to 28 L and a final pressure of 1.20 atm in two steps:(1) the gas is cooled at constant volume until its pressure has fallen to 1.20 atm, and
(2) it is heated and allowed to expand against a constant pressure of 1.20 atm until its volume reaches 28 L.
Which of the following is CORRECT?
1. w = ?6.03 kJ for the overall process
2. w = ?4.57 kJ for (1) and w = ?1.46 kJ for (2)
3. w = ?4.57 kJ for the overall process
4. w = 0 for (1) and w = ?1.46 kJ for (2)
5. w = 0 for the overall process
Explanation / Answer
1) w=0 since constant volume. 2) w = - 1.2 x 101325 x (28 - 16) x 10^-3 = -1459.08 j = -1.46 KJ 4. w = 0 for (1) and w = -1.46 kJ for (2)
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