The density (in g/L) of helium (He) gas at 723 torr and 225 K is?

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The density (in g/L) of helium (He) gas at 723 torr and 225 K is?

How many g/L ?

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Answer 1 · 2017-10-23T06:35:52

$ρ = 0.2 \cdot g \cdot L^{- 1}$ $rho=0.2*g*L^-1$

Explanation:

Now $P V = n R T$ $PV=nRT$ .....and thus $\frac{P}{R T} = \frac{n}{V} = \frac{\frac{mass}{molar mass}}{V}$ $P/(RT)=n/V=("mass"/"molar mass")/V$

And so $\frac{P}{R T} \times molar mass = \frac{mass}{V} = ρ_{helium}$ $P/(RT)xx"molar mass"="mass"/V=rho_"helium"$ as required...

ANd we assume of molar quantity of $1 \cdot m o l$ $1*mol$ under the given conditions....

$ρ_{helium} = \frac{\frac{723 \cdot Torr}{760 \cdot Torr \cdot a t m^{- 1}} \times 4.003 \cdot g \cdot m o l^{- 1}}{0.0821 \frac{L \cdot a t m}{K \cdot m o l} \times 225 \cdot K}$ $rho_"helium"=((723*"Torr")/(760*"Torr"*atm^-1)xx4.003*g*mol^-1)/(0.0821(L*atm)/(K*mol)xx225*K)$

$= 0.206 \cdot g \cdot L^{- 1}$ $=0.206*g*L^-1$

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The density (in g/L) of helium (He) gas at 723 torr and 225 K is?

How many g/L ?

1 Answer

Explanation:

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