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Chapter 18 Hm. 11

Chapter 18 Hm. 11 - Solutions to Homework#11 Chapter 18...

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Solutions to Homework #11 Chapter 18 Picture the Problem : A monatomic gas undergoes a process in which work is done on the gas, resulting in an increase in temperature. During the process heat may enter or leave the gas. Strategy: Combine the change in energy for a monatomic gas ( 29 3 f i 2 U n R T T = - with the First Law of Thermodynamics (equation 18-3) to solve for the heat flow. Solution: Solve equation 18-3 for Q : ( 29 ( 29 ( 29 ( 29 3 f i 2 3 2 560 J 4 mol 8.31 J/ mol K 130 C 5.9 kJ Q W U W n R T T = + ∆ = + - = - + × ° = Insight: It was not necessary to convert the temperature difference from Celsius degrees to kelvins in this problem because temperature differences are the same in both scales. Picture the Problem : A system that is thermally isolated from its surroundings undergoes a process in which its internal energy increases. Strategy: Use the First Law of Thermodynamics to calculate the work done during an adiabatic process ( 0 Q = ). Solution: 1. Solve the First Law for W , setting Q = 0: U Q W W U = - = -∆ 2. (a) Since the internal energy increased ( 0 U ) the work must be negative, which means it is done on the system. 3. (b) Solve numerically: ( 29 670 J 670 J W U = -∆ = - = - Insight: When no heat can enter or leave a system, any change in internal energy is equal to the work done on the system. Picture the Problem : A monatomic ideal gas expands at constant temperature. Strategy: Use the Ideal Gas Law to solve for the constant temperature. Then use equation 18-5 to solve for the work. Solution: 1. (a) Solve the Ideal Gas Law for the temperature: PV PV nRT T nR = = 2. Insert numeric values: ( 29 ( 29 3 3 i f 100 10 Pa 4.00 m 145 mol 8.31 J/ mol K 332 K T T × = = × = 3. (b) Write equation 18-5, ( 29 3 3 f f i i 3 i i 4.00 m ln ln 400 kPa 1.00 m ln 555 kJ 1.00 m V V W nRT PV V V = = = = ÷ ÷ ÷

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using the ideal gas law to replace nRT PV = : Insight: If the gas had expanded at constant pressure to the same final volume, and then cooled at constant volume to the final pressure, the total work done would have been 1200 kJ.
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