Question

1. The volume of water relative to its volume at STP at various temperatures and pressures is: T V T/C 40 60 80 1 V (1 am) 1.0074 .0115 1.0164 (500 am) 0.9867 0.9967 1.0071 Determine the expansion coefficient, a, (at 60°C for both p=1 and p=500 atm) and the isothermal compressibility, kr (at 250 atm for T-40, 60 and 80°C) a ==C), K = - ),

Answer 1

Solution for given data As for the ent specifi'e presente, kro at specific temperatun, Determine, Using Data provided expansion Constanta at 68°c. for p= 1 athu. Góc, Vat I atm = 1.0115 x Vste Yog Vat latin = 10074 x USAP Vstp = Volume of water at St Ps 224 litre *st*, en comentario) az 2.0267*16.4 °C -1 at Latm. Lofm Scanned with CamScanner

Similarly for 500 ath. x a x VSTPX/0.9967-0.98671 0.9967X Ustpl 60-40 ) scoate FX = 5.0165 x10-40-27 Now for in othermal compressibility ay at 250 atm. in found by interpolation, Kr= - for T=40% CS / Scannert. Die 4 at latu v=0.9867 for у (кР/Т 500 atun

Ky = - x (6) 280 V250 atm in calculated by interpolation, Vigo - V.00 Ne- Voo ] Keep 250 5001 - 500 Veso - 0.9867KVstp 1.0074-0.98677 Vstp 1 -499 . . -250 A 4o. Viso = 0.99707 Vstp] scanned with CamScanner

Now formula use kr = .1 X Uspp * [10074-0.9867] •99207 x USTPPL-500 ] kr = 4.1605 818-5 atm 27 at 40° Now at 60°C Vaso - Vsoo - = N-16oo - 499 (1.0115=.9967 Inste - 499 V25o.-0.9967 Vstp = Scanned with 50 CamScanner

V250 = 1,0044 X Vs op of Hyo) - Î x [1.0115-0.9967] * Vsto 1.004 1 X Uste 1 - 500 kr= ICS Scanned with camScanner

= 2.9538810-5 atm 1 at 250 caten Now at 80°c Veso - Vsoo . Ve - 1500 250 - 5001 - 500 Nosor 1.0071x Ustep [10164-6.00 74] Vstp - 250 -499 Naso = 101176 x Vstop] 1 - 500 Now at 800 Kt= 1 (V2 - V500) . Vaso ( 1-500) sou L k 11.0164-1.00717x sto - 1091768 STP 1-500 Tko = 1.84206x10-5 atm2) at 250oten Summary for Nothermal compressibility hot given the volume at 250 atm so we can find by inferpolation and then we can calculate it on same way. Please give rating and like the answer, es Scanne with Cam's canner