Question

question details in pic. answer please

1. the expansion work in kJ is ?

2. the final temperature of the ideal gas in K is?

3. the change in entropy of the gas during this process in kJ/K is?

Problems 18-20 are based on the following description An Insulated rigid tank is divided into two equal parts by a partition. Initially, one part contains 10 kg of an ideal gas with molar mass M-16 kg/kmol, a pressure of 330 kPa, and a temperature of 50 oC. The other part is evacuated. The partition is now removed and the gas fills the entire tank. (R-8.31447 kJ/kmol.K) The expansion work in KJ is closer to: 1678 725 O -356 267 820 3215 512 Question 19 The final temperature of the ideal gas in K is closer to: 323 212 273

Answer 1

18) Solution : lokg ideal gas (Vaccon 330 KPG 50°C given, : 10 The gas m, : 10kg P, : 330 KP9 ond 7, : 50°c Assumption : behoves as an ideal gas change in Kinetic and potential energy ore negligible The tank is perfecity insupored Soy the process is adiabatic, 2:0 Analysis : Since there is no restraini force during this expansion, so, the boundary work for the expansion is Also Zero. so, Expansion Work in ks is o ining

2:7 balance 191 Solution : - Assumption: • The Jainetic and potential energy changes are negligible. → The tank is insulared, thus Reak transfer is negligible. Analysis: We take the tank as our system This is a closed system and the energy could be defined as the following The 1st how for closed system is Q-Wos suf A E pot f S Ersin Here, A E bot = 0; A Elcin =0,020 mg Wb=0 JU== m (43.4) = 0 =) [U₂ = 4 fie. F=T 397, 37, : 50°C = (50+273) K son - 320K | Any Ta The Final temperature of the gas = 3236, →

os = 10kg 3) To find, the change entropy of the gas during this proces in ks/kis in ASE "(cien to y te en Asx F₂ = t; OSE NR en ( 4 KS]«mOPK) 141 Number X/8.319 16 kg/km of xen (2) IN= Max Mopar Mes => AS = 3.6 kJ/k Mole ty change in entropy of the gas of