# hw28b - aljabr(faa335 Hw28 Ross(89251 This print-out should...

This preview shows pages 1–2. Sign up to view the full content.

aljabr (faa335) – Hw28 – Ross – (89251) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points One mole of an ideal gas does 1296 J of work on the surroundings as it expands isother- mally to a final pressure of 1 atm and volume of 30 L. Determine the initial volume.( R = 8 . 31451 J / K · mol.) Correct answer: 19 . 5846 L. Explanation: Given : n = 1 mol , W = 1296 J , R = 8 . 31451 J / K · mol , P f = 1 atm , and V f = 30 L . By the ideal gas law, P V = n R T . The work done in an isothermal process is W = n R T ln parenleftbigg V f V i parenrightbigg = P V ln parenleftbigg V f V i parenrightbigg ln parenleftbigg V f V i parenrightbigg = W P V V f V i = e W/ ( P V ) V i = V f e - W/ ( P V ) Since W PV = 1296 J (1 atm) (1 . 013 × 10 5 ) (30 L) (1 × 10 - 3 ) = 0 . 426456 V i = (30 L) e ( - 0 . 426456) = 19 . 5846 L . 002 (part 2 of 2) 10.0 points Determine the temperature of the gas. Correct answer: 365 . 506 K. Explanation: According to the equation of state for an ideal gas, the temperature is T f = P f V f n R = (1 atm)(1 . 013 × 10 5 )(30 L)(1 e - 3) (1 mol)(8 . 31451 J / K · mol) = 365 . 506 K . 003 (part 1 of 3) 10.0 points Given: R = 8 . 31451 J / K · mol .

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 ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern