University:
Viterbo University
Course:
PHYS 260 | University Physics I
Academic year:

2023
Views:

23

Pages:

15
Author:

TomAnd45

shet 7 Extra problems- hon ideal gas. P=(249) 000662) W = ? If Ti: 315 k TF: 325 K P Consta W=P(Vi) at constant pressure P= 24.9 -0.00662 12 P: 24.9 TF-0-00662 T2 NE F W = P [24-9TF-0.00662 TF (24.975-0.00662 T 725 315 W = 24.9[TF-T] - 0.00662 [2] M = 206.63 J air bubble 2V-20 cm³ d=40m at the bottom of lake at Surface of water. bike at the bottom of lake Pressure on the bubble water atriospheric pressure .g.d density Pi Po + P-g-d n=(Po+Pgd vi RTI at surface water:- Pressure on bubble etwaspheric pressure" PFVFRTF n= PAVE POVE RTF RTF Case (1) closed valve frem equal V=20140 620+273 4+273 +18x981040 V = 1.036 m³ *= 1.03*16² cm² 2 closed value. V8=4VA-P8= 105 pa TB-400k P=PA open Valve TATA, TBTB PAVAARTA PBVB NBRTB = A + B n = PAVA + PVB RTA RTB VBAVA PB4 TB RTA *Case(2) opened valve AVA=ARTA PRVB MARTP PAVA RTA + RTB VB = 4VA still constat PVA PAVA PVA RTA RTA [+] 3