What is the density of neon gas given a pressure of $505 \cdot m m \cdot H g$ $505mmHg$ , and a temperature of $318 \cdot K$ $318*K$ ?

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What is the density of neon gas given a pressure of $505 \cdot m m \cdot H g$ $505mmHg$ , and a temperature of $318 \cdot K$ $318*K$ ?

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Answer 1 · 2017-03-14T04:00:18

We assume a litre volume of gas under these conditions.

And get $ρ = 0.5 \cdot g \cdot L^{- 1}$ $rho=0.5*g*L^-1$

Explanation:

Now, we know $P V = n R T$ $PV=nRT$ , for an Ideal Gas, and we ASSUME that neon will approximate Ideal Gas behaviour.

From the equation: $\frac{P}{R T} = \frac{n}{V} = \frac{\frac{mass}{molar mass}}{V}$ $P/(RT)=n/V=("mass"/"molar mass")/V$

$\frac{P}{R T} = \frac{\frac{mass}{molar mass}}{V} = ρ \times \frac{1}{molar mass}$ $P/(RT)=("mass"/"molar mass")/V=rhoxx1/"molar mass"$

And thus, (FINALLY!): $ρ = \frac{P}{R T} \times molar mass$ $rho=P/(RT)xx"molar mass"$

And we plug in the given numbers:

$ρ = \frac{\frac{505 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t m^{- 1}} \times 20.18 \cdot g \cdot m o l^{- 1}}{0.0821 \cdot \frac{L \cdot a t m}{K^{- 1} \cdot m o l^{- 1}} \times 318 \cdot K} ≅ 0.51 \cdot g \cdot L^{- 1}$ $rho=((505*mm*Hg)/(760*mm*Hg*atm^-1)xx20.18*g*mol^-1)/(0.0821*(L*atm)/(K^-1*mol^-1)xx318*K)~=0.51*g*L^-1$

The units are appropriate for density, as we require. Please check my figures and assumptions; there is a lot of 'rithmetic here.

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What is the density of neon gas given a pressure of $505 \cdot m m \cdot H g$ $505mmHg$ , and a temperature of $318 \cdot K$ $318*K$ ?