Respuesta :
ANSWERS
(a) 2 m³
(b) 2.86 kg
EXPLANATION
(a) The ideal gas equation is,
[tex]PV=nRT[/tex]Where P is the pressure of the gas, V is the volume it occupies, T is the temperature, n is the number of moles of gas in the sample and R is the universal gas constant.
We want to find the volume of a sample of oxygen if the temperature and pressure are changed. This means that we will have the same number of moles of gas, n and, since R is a constant, the product nR is constant,
[tex]\frac{PV}{T}=constant[/tex]In this case, the initial state of the gas is P₁ = 752 mmHg, V₁ = 2.17 m³, T₁ = 20°C. We have to find the volume V₂ when P₂ = 760 mmHg and T₂ = 0°C. Use the relationship,
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]Note that the temperature must be in Kelvin, not degrees Celsius. For that, add 273.15 to each temperature,
[tex]\begin{gathered} \frac{752mmHg\cdot2.17m^3_{}}{(20+273.15)K}=\frac{760mmHg\cdot V_2}{(0+273.15)K} \\ \\ \frac{1631.84}{293.15}\cdot\frac{mmHg\cdot m^3}{K}=\frac{760}{273.15}\cdot\frac{mmHg}{K}\cdot V_2 \end{gathered}[/tex]Solving for V₂,
[tex]V_2=\frac{\frac{1631.84}{293.15}\cdot\frac{mmHg\cdot m^3}{K}}{\frac{760}{273.15}\cdot\frac{mmHg}{K}}\approx2m^3[/tex]Hence, the volume of the same sample of oxygen at standard temperature and pressure is 2 m³.
(b) To find the mass of oxygen, we have to use the density of that gas,
[tex]\rho=\frac{m}{V}[/tex]We know that the density ρ = 1.43 kg/m³ and that the volume at standard temperature and pressure is 2m³. Solve the equation above for m,
[tex]m=\rho\cdot V=1.43\frac{kg}{m^3}\cdot2m^3=2.86kg[/tex]Hence, there are 2.86 kilograms of oxygen in the sample.
