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lecture 02 2010 - 6 Pressure-Volume Work Lets consider a...

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6 Pressure-Volume Work Let’s consider a simple example of the work done on the system (considered to be a gas) when a constant external pressure is applied to it. Work done on the system = -(external force × displacement) External force = external pressure × surface area W = -P × A × Δ h (note the minus sign) A × Δ h = Δ V = V final - V initial W = -P Δ V = -P(V final - V initial ) Let’s do a couple of sample problems: Apply an external pressure of 1.0 atmospheres as a gas expands from 0.5 L to 1.0 L. W = -1.0 atm (1.0 L – 0.5 L) = -0.5 L-atm. Not a particularly useful unit. Since pressure is a force per unit area, the appropriate mechanical SI unit for pressure is N/m 2 . Turns out that atmospheric pressure is 1.013 × 10 5 N/m 2 . Remember that 1 liter = 1000 cm 3 = 10 -3 m 3 . (Do you know why?). Therefore 1 L-atm = 1.013 × 10 5 N/m 2 × 10 3 m 3 = 101.3 N-m = 101.3 J This is an important unit conversion. For the expansion W = -0.5 L-atm =-50.6 J. Note that negative work is done on the system. Note also for a compression , the final volume is smaller than the initial volume. So, the work of compression on the system is positive .
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7 P V q V Δ T Absorption of Heat According to the First Law, Δ E = q + w = q - P Δ V, assuming P-V work is the only kind that can occur. Therefore, Δ E = q V . The subscript means that the process occurs at constant volume. So, if we can control our system and maintain constant volume, we know that if we inject heat into a constant volume system then we know the change in internal energy. For an ideal gas, an increase in E results in a temperature increase.
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