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Let us calculate the work done on the gas in the expansion from state i to state f. The work done on the gas is given by Equation 20.8. Because the gas is ideal and the Because T is constant in this case, it can be removed from the integral along with n To evaluate the integral, we used ∫(dx/x) ! ln x. Evaluating this at the initial and final volumes, we have (20.13) Numerically, this work W equals the negative of the shaded area under the PV curve f ( V i and the value for the work done on the gas is negative, as we expect. If the gas is compressed, then V f ' V i and the work done on the gas is positive. W ! nRT ln $ V i V f %
W ! &nRT " V f V i
dV V ! & nRT ln
V & V f V i W ! & " V f V i
P
dV ! & " V f V i
nRT V dV S E C T I O N 2 0 . 6 • Some Applications of the First Law of Thermodynamics 621 f i V PV = constant Isotherm P P i P f V i V f Figure 20.8 The PV diagram for an isothermal expansion of an ideal gas from an initial state to a final state. The curve is a hyperbola. Quick Quiz 20.6 Characterize the paths in Figure 20.9 as isobaric, isovolu- metric, isothermal, or adiabatic. Note that Q ! 0 for path B. A B C D V P T 1 T 3 T 2 T 4 Figure 20.9 (Quick Quiz 20.6) Identify the nature of paths A, B, C, and D. Example 20.6 An Isothermal Expansion A 1.0-mol sample of an ideal gas is kept at 0.0°C during an (A) How much work is done on the gas during the expan- sion? Solution Substituting the values into Equation 20.13, we W ! nRT
ln $ V i V f % ! (B) How much energy transfer by heat occurs with the surroundings in this process? & 2.7 % 10 3 J
! (1.0 mol)(8.31 J/mol$K)(273 K)
ln
$ 3.0 L 10.0 L % |