Question

2. Compute w,q, and AU for the following processes by an ideal gas: 1) irreversible expansion against a constant external pressure of 2.00 atm from 5.00 L to 10.00 L at 30°C. 2) one irreversible compression using minimum external pressure to achieve the reverse process.

Answer 1

Application of First Law of Thermodynamics---

(2 Ideal gas X Irreversible expansion at constant external pressure , Pext = 2 atm, from volume, vi=5L to volume, V₂ = 10L at constant temperature, T=30°c. we know, from 1st law of the modynamics: 0.082" AU= q + w for an ideal gasty su Now, Cu= heat na (T2-T capacity na no. of moles. o Tz = final temps Ti= Initial tempo Since, T,= Tz =(273+30) k t = 303k Hence, 2 -w Now, for Irreversible process. We-Pext (V2-V) =-2 (10-5) L.am 10 Lam = 10 x 0.314 J =-10 L.atm = 1013.9 Joules R = 0.082 Liatm-mot.at |--1013.9 Joules ↑ = 8.314 J. mollut Hence, q = - Yuniversal gas constant) = 1013.9 Joules 2) If the process is reversed, to achieve the reverse process, one has to perform same sunt of work . Therefore But , since this is a compression work (w) is positive and 1=0 AU=0 as negative. as temps is temor is constant. Hence, and this case , W= 1013.9 Joules q=-1013.9 Joules