Question

General Chemistry: Gases and the Kinetic Molecular Theory

Constants NA = 6.022 x 1023 Equalities 1 atm = 760 torr (exact) 1 atm = 760 mmHg 1 atm = 29.92 inHg 1 atm = 101325 Pa 1 atm = 14.7 psi Equations 9°F TE TC +32°F 5°C 5°C TC (TE - 32°F) 9°F TK = Tc + 273.15 R L.atm 0.08206 mol. PV = nRT A gas sample containing 1.50 moles at 25°C exerts a pressure of 7.74 psi. Some gas is added to the same container and the temperature increased to 50.°C. If the pressure had also increased to 15.5 psi, how many moles of gas were added to the container?

Answer 1

Page Date: sample gas 15.5 psl Initial Moles containing 1.50 moles pressure 7.74 psi Gas added temp increased to 50°C g pressure now How midles many е gал. were added of gas 1.50 mol (1) Temperature I-25°c = 298.15k (1) Pressure 7.74 psi I atm = 14. ipsi I atm = 760 torr 760 torr = 14.7ps1 766 fore lpst 14,7 51.7 torr Ipsi 51.7 torr then 14.7 psi 400 torr Applying ideal Oreole equation for calculating volume of container P, ng (400 torry I atm V = 1.5moly 0 : 08206 Latm a 298-15k. 760 torr & apsi RT mol k 0.526 atm y v = 36.77 L atm va 36.7 L atm 0.526 atm 69. TL CS Scalmed Volume CamScanner of container 267.77 L

Page: Date: // Gas added volume were to same semain container same New V= 69.77 liten Temperature 50°c 323.15k (12) Pressure : 1 psi = 51.7 torr 15.5 5 psi 801.3 torr 15.5 psi ideal calculating no gas equation for (n) X - R Tz P, 180g3torr x lam (69.77 L) = n2 x 0.08206 Lamm x 323.15 K 760 foor molk 2 ( 1.054 atm) (69.772) = M₂ X 26.51 L atm mol 73.5 26.5 Lalm L atm/mol 2.77 mol No. of Moles added : n ng 2.77- 1.50 1.2 Amuwes CS Scanned with CamScanner