{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapter 18 Hm. 11

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

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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 = = = = ÷ ÷ ÷
Image of page 1

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

View Full Document Right Arrow Icon
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.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern