Question 6d6af

Apr 2, 2016

For Problem 1: $\text{82 L at STP}$; for Problem 2: $\text{1124 torr}$.

Explanation:

Problem 1

I assume you are asking for the volume back on Earth at STP (1 bar and 0 °C).

We can use the Combined Gas Laws equation.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2 \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

We can rearrange the Combined Gas Laws equation to get

V_2 = V_1 × P_1/P_2 × T_2/T_1#

${P}_{1} = \text{90 atm";color(white)(mmmmmml) V_1 = "2.50 L"; T_1 = "900 °F" = "482.22 °C" = "755.37 K}$
${P}_{2} = \text{1 bar" = "0.9869 atm"; V_2 = "?";color(white)(mmll) T_2 = "0 °C" = "273.15 K}$

${V}_{2} = \text{2.50 L" × (90 color(red)(cancel(color(black)("atm"))))/(0.9869 color(red)(cancel(color(black)("atm")))) × (273.15 color(red)(cancel(color(black)("K"))))/(755.37 color(red)(cancel(color(black)("K")))) = "82 L}$

Problem 2

There are twice as many dinitrogen molecules as dioxygen molecules.

So, dinitrogen is two-thirds of the total number of molecules and generates two-thirds of the total pressure.

Dinitrogen exerts two-thirds of 1686 torr, i.e. 1124 torr.