Question

PLEASE PLEASE HELP!

Calculate the root mean square velocity and kinetic energy of CO, CO2, and SO3 at 298 K
Part A. Calculate the root mean square velocity of CO at 298 K .

Part B. Calculate the root mean square velocity of CO2 at 298 K .

Part C. Calculate the root mean square velocity of SO3 at 298 K .

Part D. Calculate the kinetic energy of CO at 298 K .

Part E. Calculate the kinetic energy of CO2 at 298 K .

Part F. Calculate the kinetic energy of SO3 at 298 K .

Part G. Which gas has the greatest velocity?

Part H. The greatest Kinetic energy?

Part i. The greatest effusion rate

Answer 1

Write the equation for root mean square velocity as follows: Produse PM... (1) Where, mis mms velocity, Ris gas constant, Tis temperature in Kelvin and Mis molar mass of substance in kg/mol. Gas constant, R = 8.314 J/ molK (A) for Co, Temperature, T = 298 K molar mass of CO, M = 28.0x10 kg/mol Substitute the corresponding values in equation (1) and calculate vom 3RT 3x8.314 J/molKx298 K Vymes 28.0 x10 kg/mol V = 515 m/s Therefore, the required solution is 515 m/s - - - - - - - - - - - - - - - - - - - - - - (B) for CO2, Temperature, T = 298 K molar mass of CO2, M = 44.0x103 kg/mol Substitute the corresponding values in equation (1) and calculate v VVM PE 3RT V = 3x8.314 J/molK 298 K V 44.0x10 kg/mol = V168925 mºis ? = 411m/s V V Therefore, the required solution is 411m/s --------------- ------ --

for SO3, Temperature, T = 298 K molar mass of SO3, M = 80.0x10 kg/mol Substitute the corresponding values in equation (1) and calculate , 3RT VVM 3x8.314 J/molKx298 K Vynu = V 80.0x10 kg/mol Vos = 92908.95 m/s2 Vm = 305 m/s Therefore, the required solution is 305 m/s -------- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - (D) Kinetic energy, KE = 5mv? .....(2) For one moles of CO, m= 28.0x10 kg ve Vroues = 515 m/s from subpart(A)) Substitute the corresponding values in equation (2) and calculate KE. KE = mu? KE = *-* 28.0x108 kgx(515 m's jo KE = 3713J Therefore, the required solution is 3713)