One mole of an ideal gas expands isothermally and reversibly from 10 dm3 to 20 dm3 at 300 K. U, q and work done in the process respectively are Given: R = 8.3 J K-1 mol-1 10 = 2.3 2 = 0.30 3 = 0.48
- 0, 21.84 kJ, -1.726 J
- 0, 21.84 kJ, 21.84 kJ
- 0, 1.718 kJ, -1.718 kJ
- 0, -17.18 kJ, 1.718 J
Answer
(3) 0, 1.718 kJ, -1.718 kJ
Solution
Isothermal process of an ideal gas U = 0
w = -2.303 nRT VfVi (here 2.303 = 10 = 2.3)
w = -2.3 1 8.3 300 2010
= -2.3 8.3 300 0.30
= -1718.1 J -1.718 kJ
U = q + w = 0
q = -w = +1.718 kJ
U, q, w = 0, 1.718 kJ, -1.718 kJ
Answer: (3)