Work done:

$\overline{){\mathbf{W}}{\mathbf{=}}{\mathbf{P}}{\mathbf{\xb7}}{\mathbf{\u2206}}{\mathbf{V}}}$

P_{12} = 3P_{0}

ΔV_{12} = V_{f} - V_{0} = V_{2} - V_{1} = 3V_{0} - V_{0} = 2V_{0}

W_{12} = P_{12}•ΔV_{12} = 3P_{0}•2V_{0} = 6P_{0}V_{0}

ΔV_{26} = V_{f} - V_{0} = V_{6} - V_{2} = 3V_{0} - 3V_{0} = 0

W_{26} = P_{26}•ΔV_{26} = 3P_{0}•0 = 0

Calculate the work *W* done by the gas during process 1→2→6→5→1.

One important use for *pV* diagrams is in calculating work. The product *p**V* has the units of Pa×m3=(N/m^{2})⋅m^{3}=N⋅m=J; in fact, the absolute value of the work done by the gas (or *on* the gas) during any process equals the area under the graph corresponding to that process on the *pV* diagram. If the gas increases in volume, it does positive work; if the volume decreases, the gas does negative work (or, in other words, work is being done *on* the gas). If the volume does not change, the work done is zero.

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