# A sample of air has a volume of 140.0 mL at 67°C . At what temperature will its volume be 50.0 mL at constant pressure?

##### 1 Answer

#### Explanation:

*Volume* and *temperature* have a **direct relationship** when pressure and number of moles are **kept constant** - this is known as Charles' Law.

What that means is that when pressure and number of moles are kept constant, **increasing** the temperature will result in an **increase** in volume.

Likewise, a **decrease** in temperature will result in a **decrease** in volume.

In your case, the volume of the gas **decreased** by a factor of about **lower** by the *same factor* of about

Mathematically, this can be written as

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

It is important to remember that the temperature of the gas **must be** expressed in *Kelvin*. The initial temperature of the gas will be

#T_1 = (273.15 + 67.0)"K" = "340.15 K"#

Rearrange the above equation to solve for

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

#T_2 = (50.0 color(red)(cancel(color(black)("mL"))))/(140.0color(red)(cancel(color(black)("mL")))) * "340.15 K"#

#T_2 = "121.48 K"#

Rounded to three sig figs, the answer will be

#T_2 = color(green)("121 K")#

If you want, you can express this in *degrees Celsius*

#T_2 = 121 - 273.15 = -152^@"C"#