Chapter 6

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

This preview shows pages 1–4. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
CHAPTER 6: THERMOCHEMISTRY 163 Now, we substitute P and Δ V into Equation (6.3) of the text to solve for w .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 33

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online