Question

On a certain winter day in Utah, the average atmospheric pressure is 656 torr. What is the molar density (in mol/L) of the air if the temperature is −29

°

C?

Answer 1

Note: Here we divide 656 by 760 to convert its unit torr into atmospheric. As pressure has been converted to atmospheric so we put the value of gas constant in atmospheric unit. And in last I divide the density by 29*10^-3 to convert back it into mole.

on the question we are given Atmospheric Pressure = 656 tone Temperature = -29°C = 244K As we know :- PV = nRT R= gas constant p = pressure v=volume n= no. of moles. T = Temperature nam m= given weight = molar weight P = density = m PV = nRT = MRT PM = m Rt = PRT PM RT As we know molar weigh of air = 29g/mol = 29 x 10 kg/mol P = 656 X 29x103 cele 8.314 x 244 X 760 1 - 700 0:002 ltr atma mał x 2448 P = 656 tatm x 29 x 103 kg/mol P = 1.25 x 103 Kg Ilir PE 25 x 10-moll its = 4.3 x 1 mol en = 4.3 x 10² mol/lta