Question

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?

Answer 1

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_2`

`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.