Question

A rigid sealed `"3.70 L"` vessel contains `"0.12 mol"` of `He(g)` and `"0.24 mol"` `Ne(g)` at `25^@ "C"`. What is the total pressure in the vessel?

Answer 1

"Rigid" containers have constant volume, and we were given that `T = "298.15 K"`. Clearly, the `"mol"`s of gas won't change in this sealed container that houses two (inert) noble gases.

So, we can use the ideal gas law for total volume, total pressure, and total `"mol"`s, assuming ideality of course, to get:

`bb(PV = nRT)`

`=> color(blue)(P) = (nRT)/V = ((n_1 + n_2)RT)/V`

`= (("0.12 + 0.24 mols")("0.082057 L"cdot"atm/mol"cdot"K")("298.15 K"))/("3.70 L")`

`=` `color(blue)("2.38 atm")`

Another way to do it is via Dalton's Law of partial pressures for an ideal two-gas mixture:

`bb(P = P_1 + P_2 + . . .)`

To find `P_1` and `P_2`:

`P_1V = n_1RT`

`P_1 = (n_1RT)/V = (("0.12 mols")("0.082057 L"cdot"atm/mol"cdot"K")("298.15 K"))/("3.70 L")`

`=` `"0.79 atm"`

Similarly:

`P_2 = (n_2RT)/V = (("0.24 mols")("0.082057 L"cdot"atm/mol"cdot"K")("298.15 K"))/("3.70 L")`

`=` `"1.59 atm"`

So that the sum is, again:

`color(blue)(P) = P_1 + P_2`

`= (n_1RT)/V + (n_2RT)/V`

`= "0.79 + 1.59 atm" = color(blue)("2.38 atm")`

(In a way, we implied Dalton's Law of partial pressures in the first method.)