# Question #7ec0a

##### 1 Answer

#### 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

#color(blue)(|bar(ul(V_1/T_1 = V_2/T_2))|)" "# , where

**Standard Temperature and Pressure** conditions are defined as a pressure of

This means that you're **decreasing** the temperature of the gas from an initial **decrease** as well.

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

#color(blue)(|bar(ul(color(white)(a/a)T["K"] = t[""^@"C"] + 273.15color(white)(a/a)))|)#

Rearrange the equation that describes Charles' Law to solve for

#V_1/T_1 = V_2/T_2 implies V_2 = T_2/T_1 * 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 = color(green)(|bar(ul(color(white)(a/a)"660 mL"color(white)(a/a)))|)#