The quantities q and w are not state properties, unlike E and H. The final values depend on how the process is done- the process done at a constant pressure is different from that done at a constant volume.

Constant volume. The change in energy in a system by the first law of thermodynamics is DE = q+w. In a constant volume process, no expansion work is done, so w_v=0. Thus, q_v = DE

Constant pressure. Here, the volume will change to maintain a constant pressure, and w_p will not equal zero. In general for expansion work, w_p = -P*DV. For gases, DV can easily be computed using the ideal gas law. Since DE = q+w, the constant pressure heat flow q_p = DE -w_p and thus q_p is not equal to q_v

Example:What is the difference in q_p and q_v when 5 grams of liquid nitrogen turn into gas at 77^oK?

Solution: The difference in q_p and q_v is the expansiom work term w_p = -P*DV. For an ideal gas, PV=nRT and the volume of the gas is far larger than the volume of the liquid, so DV = V_gas - V_liquid = V_gas = nRT/P. Thus, the expansion work is

w_p = -P*DV = -P*(nRT/P) = -nRT

w = -nRT = (5.00 g/28 g/mole * 8.31 J/mol K * 77 K) = -114J

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