# Question cc924

Apr 13, 2017

The volume of the dry gas at STP is ${\text{49 cm}}^{3}$.

#### Explanation:

The gas is saturated with water vapour, so

${p}_{\textrm{g a s}} + {p}_{\textrm{w a t e r}} = {p}_{\textrm{a t m}}$, or

${p}_{\textrm{g a s}} = {p}_{\textrm{a t m}} - {p}_{\textrm{w a t e r}} = \text{765 mmHg - 5 mmHg" = "760 mmHg}$

We can use the Combined Gas Laws to calculate the volume at STP.

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

We can rearrange this formula to get

V_2 = V_1 × p_1/p_2 × T_2/T_1 

STP is defined as 1 bar and 0 °C.

${V}_{1} = {\text{50 cm}}^{3}$
p_1 = 760 color(red)(cancel(color(black)("mmHg"))) × (1 color(red)(cancel(color(black)("atm"))))/(760 color(red)(cancel(color(black)("mmHg")))) × (101.325 color(red)(cancel(color(black)("kPa"))))/(1 color(red)(cancel(color(black)("atm")))) × "1 bar"/(100 color(red)(cancel(color(black)("kPa")))) = "1.01 bar"#
${p}_{2} = \text{1 bar}$
${T}_{2} = \textcolor{w h i t e}{l l} \text{0 °C" = "273.15 K}$
${T}_{1} = \text{10 °C" = "283.15 K}$

${V}_{2} = {\text{50 cm"^3 × (1.01 color(red)(cancel(color(black)("bar"))))/(1 color(red)(cancel(color(black)("bar")))) × (273.15 color(red)(cancel(color(black)("K"))))/(283.15 color(red)(cancel(color(black)("K")))) = "49 cm}}^{3}$