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Question Number 42128 by Necxx last updated on 18/Aug/18
$${Calculate}\:{the}\:\:{density}\:{of}\:\mathrm{1}\:{mole}\:{of} \\ $$$${oxygen}\:{at}\:{a}\:{pressure}\:{of}\:\mathrm{4}×\mathrm{10}^{\mathrm{4}} {Nm}^{−\mathrm{2}} \\ $$$${and}\:{temperature}\:{of}\:\mathrm{273}.\mathrm{2}{K}. \\ $$
Commented by Necxx last updated on 18/Aug/18
$${please}\:{check}\:{this}\:{approach} \\ $$$${n}=\mathrm{1}\:{P}=\mathrm{4}×\mathrm{10}^{\mathrm{4}} {N}/{m}^{\mathrm{2}} \:{T}=\mathrm{273}.\mathrm{2}{K} \\ $$$${R}=\mathrm{8}.\mathrm{31}{Jmol}^{−\mathrm{1}} {K}^{−\mathrm{1}} \: \\ $$$${molar}\:{mass}\:{of}\:{oxygen}=\mathrm{32}×\mathrm{10}^{−\mathrm{3}} {kg}/{mol} \\ $$$${mass}\:{of}\:\mathrm{1}\:{mole}=\mathrm{3}.\mathrm{5}×\mathrm{10}^{−\mathrm{26}} \\ $$$$\therefore{V}=\frac{{nRT}}{{P}}=\frac{\mathrm{1}×\mathrm{8}.\mathrm{31}×\mathrm{273}.\mathrm{2}}{\mathrm{4}×\mathrm{10}^{\mathrm{4}} }=\mathrm{5}.\mathrm{67}×\mathrm{10}^{−\mathrm{2}} {l} \\ $$$${since}\:{density}=\rho={m}/{V}\:{then} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{3}.\mathrm{5}×\mathrm{10}^{−\mathrm{26}} }{\mathrm{5}.\mathrm{67}×\mathrm{10}^{−\mathrm{2}} }=\mathrm{0}.\mathrm{617}×\mathrm{10}^{−\mathrm{24}} {kg}/{l} \\ $$$$ \\ $$$${pls}\:{check}\:{and}\:{comment} \\ $$$$ \\ $$
Commented by Necxx last updated on 18/Aug/18
$${please}\:{help}\:{its}\:{urgent} \\ $$