Chapter%206

Chapter%206 - 6.10 If the heater operates as rated, then...

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6.10 If the heater operates as rated, then the total amount of heat transferred to the cylinder will be 1 1 4 (100 J s )(10 min)(60 s min ) 6.0 10 J or 60 kJ = × Work will be given by ext = − Δ w P V in this case because it is an expansion against a constant opposing pressure: (0.975 atm)(10.00 L 2.00 L) 7.80 L atm = − = − w In order to combine the two terms to get the internal energy change, we must first convert the units on the energy terms to the same values. A handy trick to get the conversion factor for L atm to J or vice versa is to make use of the equivalence of the ideal gas constant R in terms of L atm and J: 1 1 1 1 8.314 J K mol 7.80 L atm 790 J or 0.790 kJ 0.082 06 L atm K mol = − = − w The internal energy change is then the sum of these two numbers: 60 kJ ( 0.790 kJ) 59 kJ U q w Δ = + = + − The energy change due to the work term turns out to be negligible in this problem. 6.14 (a) true only if q = 0; (b) always true, volume is fixed and no work can be done; (c) never true; (d) always true; (e) true if q = 0 6.24 (a) Because the process is isothermal, 0 Δ = U and . = − q w For a reversible process, 2 1 ln = − V w nRT V n is obtained from the ideal gas law: 1 1 1 1 (2.57 atm)(3.42 L) 0.359 mol (0.082 06 L atm K mol )(298 K) 7.39 (0.359 mol)(8.314 J K mol )(298 K) ln 685 J 3.42 685 J = = = = − = − = + PV n RT w q (b) For step 1, because the volume is constant, 0 and . = Δ = w U q
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In step 2, there is an irreversible expansion against a constant opposing pressure, which is calculated from ex = − Δ w P V The constant opposing pressure is given, and Δ V can be obtained from final initial 7.39 L 3.42 L
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Chapter%206 - 6.10 If the heater operates as rated, then...

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