SQSM-03

# SQSM-03 - Chapter 3 First Law of Thermodynamics 1 3 What...

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Chapter 3 First Law of Thermodynamics 3-1 1. What state functions correspond to the heat absorbed at constant P and the heat absorbed at constant V? H is the heat absorbed at constant pressure. E is the heat absorbed at a constant volume. 3. Classify each of the following processes as endothermic or exothermic: a) melting a solid b) combustion of butane c) condensing a liquid endothermic exothermic exothermic d) photosynthesis e) a battery reaction endothermic exothermic 5. What is the energy change of the system if the system: a) absorbs 50. J of heat and does 50. J of work? System absorbs heat, so q = +50 J. The system does work, so w = -50 J, E = q + w = 50 J + (-50 J) = 0 J b) releases 20. J of heat and has 415 J of work done on it? The system releases heat, so q = -20 J. Work is done on the system, so w = + 415 J. E = q + w = -20 J + 415 J = 395 J 7. What is E of a gas that gives off 312 J of heat while being compressed 862 ml by a pressure of 1.64 atm? What are E sur and E univ for the process? The system is compressed, so V is negative and w = - P op V = -(1.64 atm)(-0.862 L) = 1.41 L . atm. Convert L . atm to joules: w = (1.41 L-atm)(101.3 J . L-1 . atm-1) = 143 J. Heat is given off, so q = -312 J. Use Equation 3.5 and the values of q and w : E = q + w = -312 +143 = -169 J Therefore, the energy of the system decreases by 169 J . As shown in Equation 3.4, E surr = - E = 169 J E univ = 0 (The First Law of Thermodynamics, Eq. 3.3) 9. Determine the mass of octane (C 8 H 18 , H comb = -5500. kJ/mol) must be combusted to yield the amount of energy equivalent to: First, convert the heat of combustion for octane to kJ/g: 81 8 8 8 8 1 mol C H 5500. kJ kJ 48.15 mol C H 114.23 g C H g C H ×= then, use the above to convert from the given energy into the mass of octane. a) 3.1 kJ, the kinetic energy of a 220. lb. linebacker running at a speed of 40. yd in 4.7 seconds. 8 8 1 g C H 3.1 kJ 0.064 g C H 48.15 kJ b) 17 J, the potential energy of a 5 lb bag of sugar on top of a 30. inch high counter, 17 J. 4 8 8 1 g C H 1 kJ 17 J 3.5 10 g C H 1000 J 48.15 kJ ×× = × c) the heat required to raise the temperature of 1 quart of water from 25 o C to its boiling point, 320 kJ. 8 8 1 g C H 320. kJ 6.65 g C H 48.15 kJ 11. The heat of vaporization of CS 2 at its normal boiling point of 46 o C is 26.7 kJ/mole. Calculate H for the condensing 41.2 g of CS 2 gas to liquid at 46 o C. Condensation is the reverse of vaporization, so the enthalpy of condensation has the same magnitude but is opposite in sign to the heat of vaporization. Consequently, H cond = -26.7 kJ/mol. We simply need to multiply the moles of CS 2 by the heat of condensation to obtain the H for the process. 2 22 2 1 mol CS 41.2 g CS =0.541 mol CS 76.143 g CS × Convert the moles to units of kJ using the given heat of condensation: 2 2 -26.7 kJ 0.541 mol CS =-14.4 kJ mol CS ×

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First Law of Thermodynamics 3-2 13. Use the information in the preceding exercise to answer the following: a) What mass of ice could be melted at 0 o C by 35.0 kJ of heat?
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## This note was uploaded on 01/27/2011 for the course CHEM 201 taught by Professor Wilson during the Spring '10 term at N.C. Central.

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SQSM-03 - Chapter 3 First Law of Thermodynamics 1 3 What...

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