# Chapter 6 Answers - Chemistry 141 Lecture Dr Ben Tovrog...

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Chemistry 141 Lecture Dr. Ben Tovrog Chapter 6 Supplemental Problems - Answers 1. Calculate Δ E for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system. Δ E = q + w Where q = +15.6 kJ, since ht process is endothermic, and w = +1.4 kJ, since work is done on the system. Δ E = 15.6 kJ + 1.4 kJ = 17.0 kJ. The system has gained 17.0 kJ. 2. Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm. For a gas at constant pressure, w = -P Δ V. In this case P = 15 atm, and V = 64 - 46 = 18 L. w = - 15 atm * 18 L = -270 L atm. (Note that since the gas expands, it does work on its surroundings. Energy flows out of the gas, so w is a negative quantity). 3. A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes from 4.00 x 10 6 L to 4.50 x 10 6 L by the addition of 1.3 x 10 8 J of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate E for the process (To convert between L atm and J use 1 L atm = 101.3 J) Δ E = q + w. q = +1.3 x 10 8 J. w = -P Δ V. P = 1 atm and Δ V = V initial – V final = 4.5 x 10 6 L - 4.00 x 10 6 L = 5.0 x 10 5 L. w = -1.0 atm * 5.0 x 10 5 L = - 5.0 x 10 5 L atm. (the negative sign is because the gas is expanding and doing work on the surroundings). Convert to J w = -5.0 x 10 5 L atm * (101.3 J / L atm) = -5.1 x 10 7 J. Δ E = q + w = (+1.3 x 10 8 J) + (-5.1 x 10 7 J) = 8 x 10 7 J. Since more energy is added through heating than the gas expends doing work, there is a net increase in the internal energy of the gas in the balloon. Hence Δ E is positive. 4.

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Chapter 6 Answers - Chemistry 141 Lecture Dr Ben Tovrog...

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