2atm 20°c temperature 10g O2 volume 3.76 R value?

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Answer 1 · 2018-03-05T14:43:55

$R=0.082 \ "L atm K"^-1 \ "mol"^-1$

According to the Ideal Gas Law:

$PV=nRT$

This asks us to solve for $R$ :

$R=(PV)/(nT)$

Here,

$P=2 \ "atm"$

$V=3.76 \ "L"$

$n=10/15.999=0.312 \ "mol"$

$T=293.15 \ "K"$

Inputting:

$R=(2*3.76)/(0.312*293.15)$

$R=0.082 \ "L atm K"^-1 \ "mol"^-1$