{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

image002 - 6.11(a Because the process is isothennal 310 = t...

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

View Full Document Right Arrow Icon
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 6.11 (a) Because the process is isothennal, 310' = t} and q = u-w. For n reversible process, is z —nRT In E ii I: is obtained from the ideal gas law: F? (2.5? ath3.42 L) =w = -—---—-—-—————-——-—--- = «H.359 mol " er (use: as L-atm-K" -mol"]{293 K} w = {[1359 mol}(ii.314 1-K" -rnol"'){293 K) log—:3 = 4585 I q = + 685 I {b} For step 1, because the volume is constant, to = i] and 3H! = g. In step 2, there is an irreversible expansion against a constant opposing pressure, which is calculated from w = dPnfi V The constant opposing pressure is given, and 111’ can be obtained from rm — Km, = 139 L — 3.42 L w :—(L19stm}(?.39 L —3.42 L} “H.325 I =—4.T2L-atm=(—4.?2L*atm} =4?“ iL-atm The total work forpartB is 0.1+ {-479 J} =47?! {gall No,fl1egassesdonothavethesamefinaltempershne.1heflfl4molemfles havemore internal vibrations inwhichto stmeehergythandobig molecules. Assruulgitwiflrequiremoteenerytoinereasethe temperature ofCH4[g}, andif me some amount Izlii'e-ttiergjir is supplied to bothgassesfisetemperamreoffizwfllbehigherthanflmtofflm. ...
View Full Document

{[ snackBarMessage ]}