Question

physic
3. 2 moles of an ideal gas at 17°C has a pressure of 760mm mercury, and is compressed once isothermally and then adiabatically until its volume is halved in each case reversibly and from identical initial conditions). The gas constant is 8.314J/kg. The density of the mercury is 13.60g/cm? [Express all your answers in MKS units e.g. volume in cubic meter, pressure in Pascal, Temperature in Kelvin, etc.) (a) Express the pressure of the gas in units of Pascal. P = For parts (b) through (h) assume the initial pressure of the gas is 1.013 x 10 Pa (b) What is the initial volume of the gas? Vi = Consider the isothermal process: (c) What is the final pressure of the gas?

Answer 1

ns2 T-17°C2734175 290k P = 7.60mm Mercury : 100 13x105 Pa Do Jalmis isolhearyal Temperature constant as Pů = 1.013xios Pa (6) Vi = ? PV=nRT ve - 2x8.314 x 290 Tos - 0.0482 m3 1.013X 105 T=conolant, volume halled. & a Pivi = P j . 1.0138x105 x V = Pq 9 = 2.026x105 Pa (d) for isolhermal process W= nRT la V2 = Pivi In 1/2 = nRT in = 2*8.314x290 luţ - 3342, 44 J cSScanned with CamScanner

Now By for isoluermal process au=0 hol law of luemo dynamics До - до 4 дю AQ - AWF- 3342.44 3 Aus CONSIDERING ADIABETIC PROCESS pyo = conor. • Nothing is MENTIONED 50 colISIDERING MONCATOMIC GAS Pi = 1:013x105 pa v; = v m3 Nao 2 |-01 2 10 4 2 - ( 13 x 1.013X10S =P = 3.22X10% far PV = Conal- Rva Ganesh idealgas PV = nRT MORT VP = const o TV . - Const. CSScanned with CamScanner

- Comt PV=nRT [ideal gars] V-net - P.InRTY - Const pop pop cornet - Cono 0 ) 1 Ti: 17° +17+273 52901, * -(0) 20 O13x10 322 (2.162)93 TJ = 428.8k Ang To = 428.8% -273 & 155.8°C Aus CScanned with CamScanner

3 h Adiabetic means AQ = 0 By Jat law gj you modynamics 016 conomice AQ - Autaw 4W = -AU = - nCWAT cv for = -2.3R ( 45 ) ESR monoclomis = - 3R ( 428.8–290) = - 348.314 x 158.8 --3960.78 ] Blů isolhermal nomic gas e adiabetic monoalomic csscanned with CamScanner