# 20.0 g each of helium and an unknown diatomic gas are combined in a 1500. ml container. lf the temperature is 298 K, and the pressure inside is 86.11 atm, what is the unknown gas?

Dec 15, 2016

We assume (possibly unreasonably) $\text{ideality}$, and get......$C {l}_{2}$, $\text{molecular chlorine}$, as the unknown gas..........

#### Explanation:

$P = \frac{n R T}{V}$........

And solve for $n = \frac{P V}{R T}$

$=$ $\frac{86.11 \cdot a t m \times 1.500 \cdot L}{0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1 \times 298 \cdot K}$

$n = 5.28 \cdot m o l$

But $n = {n}_{\text{He"+n_"unknown gas}}$,

i.e. $5.28 \cdot m o l = \frac{20 \cdot g}{4 \cdot g \cdot m o {l}^{-} 1} + {n}_{\text{unknown gas}}$

And thus ${n}_{\text{unknown gas}} = \left(5.28 - 5\right) \cdot m o l = 0.280 \cdot m o l$.

And thus $\text{molecular mass of unknown gas}$ $=$ $\frac{20 \cdot g}{0.280 \cdot m o l} = 71.5 \cdot g \cdot m o {l}^{-} 1$.

And clearly, since the gas is diatomic, i.e. ${X}_{2}$, $X$ has an atomic mass of $\cong 35.5 \cdot g \cdot m o {l}^{-} 1$, and thus (finally!) $X = C l$, which certainly forms a $C {l}_{2}$ molecule as required.