Chapter 6

Chapter 6 - CHAPTER 6 THERMOCHEMISTRY 6.15 Recall that the...

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CHAPTER 6 THERMOCHEMISTRY 6.15 Recall that the work in gas expansion is equal to the product of the external, opposing pressure and the change in volume. (a) w = P Δ V w = (0)(5.4 1.6)L = 0 (b) w = P Δ V w = (0.80 atm)(5.4 1.6)L = 3.0 L atm To convert the answer to joules, we write 101.3 J 3.0 L atm 1L atm =− × = 2 3.0 10 J −× w (c) w = P Δ V w = (3.7 atm)(5.4 1.6)L = 14 L atm To convert the answer to joules, we write 101.3 J 14 L atm × = 3 1.4 10 J w 6.16 (a) Because the external pressure is zero, no work is done in the expansion. w = P Δ V = (0)(89.3 26.7)mL w = 0 (b) The external, opposing pressure is 1.5 atm, so w = P Δ V = (1.5 atm)(89.3 26.7)mL 0.001 L 94 mL atm 0.094 L atm 1mL × w To convert the answer to joules, we write: 101.3 J 0.094 L atm × = 9.5 J w (c) The external, opposing pressure is 2.8 atm, so w = P Δ V = (2.8 atm)(89.3 26.7)mL 2 0.001 L ( 1.8 10 mL atm) 0.18 L atm =− × × w
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CHAPTER 6: THERMOCHEMISTRY 162 To convert the answer to joules, we write: 101.3 J 0.18 L atm 1L atm =− × = 18 J w 6.17 An expansion implies an increase in volume, therefore w must be 325 J (see the defining equation for pressure-volume work.) If the system absorbs heat, q must be + 127 J. The change in energy (internal energy) is: Δ E = q + w = 127 J 325 J = 198 J 6.18 Strategy: Compression is work done on the gas, so what is the sign for w ? Heat is released by the gas to the surroundings. Is this an endothermic or exothermic process? What is the sign for q ? Solution: To calculate the energy change of the gas ( Δ E ), we need Equation (6.1) of the text. Work of compression is positive and because heat is given off by the gas, q is negative. Therefore, we have: Δ E = q + w = 26 kJ + 74 kJ = 48 kJ As a result, the energy of the gas increases by 48 kJ. 6.19 We first find the number of moles of hydrogen gas formed in the reaction: 2 2 1molH 1molSn 50.0 g Sn 0.421 mol H 118.7 g Sn 1 mol Sn ×× = The next step is to find the volume occupied by the hydrogen gas under the given conditions. This is the change in volume. 2 (0.421 mol)(0.0821 L atm / K mol)(298 K) 10.3 L H 1.00 atm ⋅⋅ == = nRT V P The pressure-volume work done is then: 101.3 J (1.00 atm)(10.3 L) 10.3 L atm =− Δ =− × = 3 1.04 10 J PV w −× 6.20 Strategy: The work done in gas expansion is equal to the product of the external, opposing pressure and the change in volume. w = P Δ V We assume that the volume of liquid water is zero compared to that of steam. How do we calculate the volume of the steam? What is the conversion factor between L atm and J? Solution: First, we need to calculate the volume that the water vapor will occupy ( V f ). Using the ideal gas equation: 2 2 HO La tm (1 mol) 0.0821 (373 K) mol K =3 1 L (1.0 atm) ⎛⎞ ⎜⎟ ⎝⎠ nR T V P It is given that the volume occupied by liquid water is negligible. Therefore, Δ V = V f V i = 31 L 0 L = 31 L
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CHAPTER 6: THERMOCHEMISTRY 163 Now, we substitute P and Δ V into Equation (6.3) of the text to solve for w .
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Chapter 6 - CHAPTER 6 THERMOCHEMISTRY 6.15 Recall that the...

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