P21_020 - 20. (a) The final pressure is pf = (5.00...

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: 20. (a) The final pressure is pf = (5.00 kPa)e(Vi −Vf )/a = (5.00 kPa)e(1.00 m 3 −2.00 m3 )/1.00 m3 = 1.84 kPa . (b) We use the ratio form of the gas law (see Sample Problem 20-1) to find the final temperature of the gas: p f Vf (1.84 kPa)(2.00 m3 ) Tf = Ti = 441 K . = (600 K) p i Vi (5.00 kPa)(1.00 m3 ) For later purposes, we note that this result can be written “exactly” as Tf = Ti 2e−1 . In our solution, we are avoiding using the “one mole” datum since it is not clear how precise it is. (c) The work done by the gas is f W Vf p dV = = (5.00 kPa)e(Vi −V )/a dV Vi i Vf = (5.00 kPa)eVi /a · − ae−V /a = = (5.00 kPa)e1.00 (1.00 m3 ) e−1.00 − e−2.00 3.16 kJ . Vi (d) Consideration of a two-stage process as suggested in the hint, brings us simply to Eq. 21-4. Consequently, with CV = 3 R (see Eq. 20-43), we find 2 ∆S Vf Vi = nR ln = nR ln 2 + = = = +n Tf 3 R ln 2 Ti 3 ln 2e−1 2 3 3 p i Vi ln 2 + ln 2 + ln e−1 Ti 2 2 3 (5000 Pa)(1.00 m ) 5 3 ln 2 − 600 K 2 2 1.94 J/K . ...
View Full Document

This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

Ask a homework question - tutors are online