Question

When 2 g of gaseous substance A are introduced into an initially evacuated flask kept at 25°C, the pressure is found to be 1 atm. Three grams of gaseous substance B are then added to the 2g of A, and the pressure is now found to be 1.5 atm. Assuming ideal gas behavior. Calculating the ratio of molecular weights, that is, MwA /MwB

Answer 1

1/3

Step-by-step explanation:

From the ideal gases law we have:

`PV=nRT`

P= pressure, V= volume of the container, n= mol of gas, R=constant of gases and T=temperature.

The molecular weight is:

`M_(w)=(m)/(n)`

m is mass.

The dalton's law tell us that the total pressure of a mixture of gases is the sum of the partial pressure of every gas.

`P_(T)=P_(A)+P_(B)`

the relationship between the molecular weights is:

`(M_(wA))/(M_(wB))=(m_(A)*n_(B))/(m_(B)*n_(A))`(equation 1)

where

`n_(A)=(P_(A)V)/(RT)`

`n_(B)=(P_(B)V)/(RT)`

Replacing in equation 1

`(M_(wA))/(M_(wB))=(m_(A))/(m_(B))*(P_(B)VRT)/(P_(A)VRT)`

since V and T are constante

`(M_(wA))/(M_(wB))=(m_(A)*P_(B))/(m_(B)*P_(A))`

we don't know the partial pressure of B but we know the partial pressure of A since it was the pressure of A at the beginning before adding B (1 atm). We also also know the total pressure of the container with both gases (1.5 atm). From the dalton's law

`P_(B)=P_(T)-P_(A) - - -> 1.5 atm - 1 atm = 0.5 atm`

now

`(M_(wA))/(M_(wB))=(2g*0.5atm)/(3g*1atm)=(1)/(3)`