# Question #36619

##### 1 Answer

Because its **absolute temperature** didn't double.

#### Explanation:

As you know, when the pressure and the number of moles of gas present in a container are kept **constant**, the volume of the gas and its temperature have a **direct relationship** as described by **Charles' Law**.

#V_1/T_1 = V_2/T_2#

Here

#V_1# ,#T_1# represent the volume and theabsolute temperatureof the gas at an initial state#V_2# ,#T_2# represent the volume and theabsolute temperatureof the gas at a final state

The key thing to keep in mind here is that the temperature of the gas **must** be expressed in Kelvin in order for this equation to work.

To do that, you use the conversion factor

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

In your case, the temperature of the gas doubles in degrees Celsius, but that does not mean that it will double in Kelvin!

#T_1 = 25^@"C" + 273.15 = "298 K"#

#T_2 = 50^@"C" + 273.15 = "323 K"#

This means that the volume of the gas, which is equal to

#V_2 = T_2/T_1 * V_1#

will be equal to

#V_2 = (323 color(red)(cancel(color(black)("K"))))/(298color(red)(cancel(color(black)("K")))) * V_1#

#V_2 ~~ 1.084 * V_1#

So the volume of the gas will only increase **times** when the temperature of the gas goes from

In order for the volume to **double**, i.e. to have

#T_2 = (2 * color(red)(cancel(color(black)(V_1))))/color(red)(cancel(color(black)(V_1))) * T_1#

#T_2 = 2 * T_1#

#T_2 = "596 K"#

In degrees Celsius, the temperature of the gas must increase to

#t_2 = "596 K"- 273.15 = 323^@"C"#

in order for the volume of the gas to **double**.