Question

A gas at 362 K occupies a volume of 0.67 L. At what temperature will the volume increase to 1.12 L?

Answer 1

The temperature at which the volume increases to `"1.12 L"` is `6.1xx10^2` `color(white)(.)"K"`.

Explanation:

This is an example of Charles' law , which states that the volume of a gas, held at constant pressure and amount, varies directly with the temperature in Kelvins. This means that when the volume increases, the temperature increases and vice-versa.

The equation to use is:

`V_1/T_1=V_2/T_2`

where `V_1` is the initial volume, `T_1` is the initial temperature, `V_2` is the final volume, and `"T_2"` is the final temperature.

Given/Known
`T_1="362 K"`
`V_1="0.67 L"`
`V_2="1.12 L"`

Unknown: `T_2`

Solution
Rearrange the equation to isolate `T_2`, then substitute the known values into the equation and solve.

`T_2=(V_2T_1)/V_1`

`T_2=(1.12"L"xx362"K")/(0.67"L")=6.1xx10^2` `color(white)(.) "K"`
(`"605 K"` rounded to two significant figures)