work = ∆q - ∆(PV)
∆q = 0 (isothermal process)
- ∆(PV) = -[P∆V + V∆P]
P∆V = 0 since V = constant
work = - V∆P
∆P = (2 atm -1 atm) = 1 atm
∆P = 101.325 kPa
work = -(1 dm^3)(101.325 kPa)
work = 101.325J
1 mol of an ideal gas expands under isothermal conditions (T=298.15K). The initial pressure and volume is 2.00 atm and 1.00L respectively, and the final pressure is 1.00 atm. Find the expansion work. The answer is w = -101.325 J. I am unsure how to get this answer. Can someone please walk me through the steps. Thanks.
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