The most common form of work that we deal with in chemistry is expansion work, work caused by the expansion or contraction of the system. This occurs in any chemical reaction that absorbs or generates gases and is done at a constant pressure. (Keep in mind that our definition of enthalpy is constant pressure heat flow.)

Computing the work is easy: expansion work is defined to be the product of the force on the system times the distance through which the force moves. The negative sign is due to the fact that expansion work is done on the system by the surroundings. (If you want to compute the work done on the surroudings by the system, remove the - sign.)

w = -f*d

If we do the reaction in a piston where the piston moves up and down in response to pressure changes, then the force is just the pressure times the surface area of the piston, w = -P*a*d. In the same vein, for any expansion against a constant external pressure is just the change in volume times the pressure, w = -P*DV

Example: What is the work done on the system by the surroundings when 1 gram of liquid nitrogen converts to nitrogen gas at 77^oK at a constant pressure of 1 atm? What is the work done by the system on the surroundings?

Solution: The expansion work is w = -P*DV. The change in volume is V_gas - V_liquid. For liquid nitrogen, the volume is much less than the volume of the gas so we can assume that it is zero.

The volume of the gas at that temperature can be found by the ideal gas law: PV =nRT, V = nRT/P. Plug this into the above work equation and we find that

w = -P*DV = -P*V_gas = -P*(nRT/P) = nRT

w = -(1 g/28 g/mole * 8.31 J/mol K * 77 K) = 22.9 J.

The work done on the surroundings by the system is thus -22.9 J

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