# The temperature of 3.0L of a gas is changed from 27° C to -10° C at constant pressure. What is the new volume in mL?

##### 1 Answer

Here's what I got.

#### Explanation:

Your tool of choice here is the equation that describes **Charles' Law**

#color(blue)(ul(color(black)(V_1/T_1 = V_2/T_2)))#

Here

#V_1# and#T_1# represent the volume and the absolute temperature of the gas at an initial state#V_2# and#T_2# represent the volume and the absolute temperature of the gas at a final state

Charles Law describes the **direct relationship** that exists between the volume and the temperature of a gas when the pressure and the number of moles of gas are kept constant.

Simply put, when these conditions are met, *increasing* the temperature of the gas will cause its volume to **increase**. Similarly, *decreasing* the temperature of the gas will cause its volume to **decrease**.

In your case, the temperature is *decreasing*, so you should expect to have

#V_2 " " < " " "3.0 L"#

Rearrange the equation to solve for

#V_1/T_1 = V_2/T_2 implies V_2 = T_2/T_1 * V_1#

Before plugging in your values, make sure to convert the temperatures of the gas from *degrees Celsius* to *Kelvin* by using the fact that

#color(blue)(ul(color(black)(T["K"] = t[""^@"C"] + 273.15)))#

You will end up with

#V_2 = ((-10 + 273.15)color(red)(cancel(color(black)("K"))))/((27 + 273.15)color(red)(cancel(color(black)("K")))) * "3.0 L" = "2.6 L"#

Expressed in *milliliters*, the answer will be

#color(darkgreen)(ul(color(black)(V_2 = 2.6 * 10^3color(white)(.)"mL")))#

I'll leave the answer rounded to two **sig figs**.