The ideal gas law

Tutorial 4

                

The mass density of oxygen is $ \rho_2 = 1.43 \;\frac{g}{L} $ under normal conditions (ntp: $ 0.00 \;^oC $; $ 1,00 \;atm $) . What is this density $ \rho_1 $ in NTP conditions ($ 20.00^oC $; $ 1.00\; bar $)? Solution (Click): Conversion to Kelvin: ..........$T_2=$ $273.15\; K$ ; $T_1=$ $293.15\; K$ Conversion to atm: ..........$P_1$ = $\frac{1}{1.0133}\;atm$ Conditions 2: $1,43\;g$ dans $1,00\;L$ Conditions 1: $1,43\;g$ dans $\color{red}?$ $\;L$ Ideal gas law: ..........$\frac{P_1\cdot V_1 }{T_1}=\frac{P_2\cdot V_2 }{T_2}$ Calculation: ..........$V_1=\frac{T_1\cdot V_2\cdot P_2}{T_2\cdot P_1}$ = $\frac{293.15 \cdot 1.00 \cdot 1.00}{273.15\cdot \frac{1}{1.0133}} = 1.09 \;L$ Answer: ..........$\rho_1=\frac{m_1}{V_1}$ = $\frac{1.43}{1.09} = 1.31 \;\frac{g}{L}$