# 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?

Dec 29, 2016

"Rigid" containers have constant volume, and we were given that $T = \text{298.15 K}$. Clearly, the $\text{mol}$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 $\text{mol}$s, assuming ideality of course, to get:

$\boldsymbol{P V = n R T}$

$\implies \textcolor{b l u e}{P} = \frac{n R T}{V} = \frac{\left({n}_{1} + {n}_{2}\right) R T}{V}$

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

$=$ $\textcolor{b l u e}{\text{2.38 atm}}$

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

$\boldsymbol{P = {P}_{1} + {P}_{2} + . . .}$

To find ${P}_{1}$ and ${P}_{2}$:

${P}_{1} V = {n}_{1} R T$

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

$=$ $\text{0.79 atm}$

Similarly:

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

$=$ $\text{1.59 atm}$

So that the sum is, again:

$\textcolor{b l u e}{P} = {P}_{1} + {P}_{2}$

$= \frac{{n}_{1} R T}{V} + \frac{{n}_{2} R T}{V}$

= "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.)