# Question 7ec0a

Mar 6, 2016

$\text{660 mL}$

#### Explanation:

You know that the pressure and the number of moles of gas are kept constant, which means that you can use Charles' Law to find the volume of the gas at standard temperature.

Charles' Law states that the volume and the temperature of a gas have a direct relationship when pressure and number of moles of gas are kept constant.

Simply put, if you increase the temperature of a gas, its volume increases as well. Likewise, if you decrease the temperature of a gas, its volume decreases as well.

Mathematically, this is written as

$\textcolor{b l u e}{| \overline{\underline{{V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}}} |} \text{ }$, where

${V}_{1}$, ${T}_{1}$ - the volume and absolute temperature of the gas at an initial state
${V}_{2}$, ${T}_{2}$ - the volume and temperature of the gas at a final state

Standard Temperature and Pressure conditions are defined as a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$.

This means that you're decreasing the temperature of the gas from an initial ${25}^{\circ} \text{C}$ to ${0}^{\circ} \text{C}$. As a result, you should expect the volume to decrease as well.

To convert between degrees Celsius and Kelvin, use the conversion factor

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} T \left[\text{K"] = t[""^@"C}\right] + 273.15 \textcolor{w h i t e}{\frac{a}{a}}}} |}$

Rearrange the equation that describes Charles' Law to solve for ${V}_{2}$

${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2} \implies {V}_{2} = {T}_{2} / {T}_{1} \cdot {V}_{1}$

Plug in your values to get

V_2 = ((273.15 + 0)color(red)(cancel(color(black)("K"))))/((273.15 + 25)color(red)(cancel(color(black)("K")))) * "725 mL" = "664.21 mL"#

Rounded to two sig figs, the number of sig figs you have for the initial temperature of the gas, the answer will be

${V}_{2} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{660 mL} \textcolor{w h i t e}{\frac{a}{a}}}} |}$