Question

When 2g of gas A are introduced into a flask at 25C the pressure is 1atm. 3g of gas B are added, the temperature remained the same but the pressure is 1.5atm. Assume ideal gas behavior and compute the ratio of molecular weights MA/MB

Answer 1

Volume and Temperatures are constant

we have PV = nRT , when V , T are constanst then   P/n = constant , hence P2/P1 = n2/n1

we see that P1 = 1 , P2 = 1.5 , additional pressure P2-P1 is due to B moles added

moles of A = mass of A / molar mass of A = 2/MA ,

moles of B = 3/MB

P1 = 1 = (2/MA) x RT / V   ...............(1)

P2-P1 = 0.5 = ( 3/MB) x RT/V ..............(2)

now (2) / ( 1) gives   ( 0.5/1) = ( MA/MB) x ( 3/2)

MA/MB = 0.333