A gas thermometer contains 250 mL of gas at 0° C and 1.0 atm pressure. If the pressure remains at 1.0 atm, how many mL will the volume increase for every one celsius degree that the temperature rises?

I don't understand this part "for every one celsius degree"

1 Answer
Aug 15, 2017

Here's what I got.

Explanation:

The problem states that the pressure remains unchanged, so right from the start, you should know that you can use Charles' Law to calculate the change in volume associated with a "1-"""^@"C"1-C increase in the temperature of the gas.

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

Now, it's very important to remember that you must work with absolute temperatures here, i.e. with temperatures expressed in Kelvin.

In your case, the gas starts at

0^@"C" = 0^@"C" + 273.15 = "273.15 K"

When the temperature of the gas increases by 1^@"C", it also increases by "1 K" because you have

0^@"C" + 1^@"C" = 1^@"C" -> the temperatue increases by 1^@"C"

which is

1^@"C" = 1^@"C" + 273.15 = "274.15 K"

Similarly, you will have

"273.15 K" + "1 K" = "274.15 K" -> the temperature increases by "1 K"

So, you must determine the change in volume that accompanies a 1^@"C" = "1 K" increase in temperature. If you start at 0^@"C" = "273.15 K", you will end up at "274.15 K".

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

Plug in your values to find

V_2 = (274.15 color(red)(cancel(color(black)("K"))))/(273.15color(red)(cancel(color(black)("K")))) * "250 mL" = "250.9 mL"

So the volume of the gas increased by

overbrace("250.9 mL")^(color(blue)("at 274.15 K" = 1^@"C")) - overbrace("250 mL")^(color(blue)("at 273.15 K" = 0^@"C")) = "0.9 mL"

Notice what happens when you increase the temperature from "274.15 K" to "275.15 K". You will once again have--remember to use the volume of the gas at "274.15 mL" as V_1

V_2 = (275.15 color(red)(cancel(color(black)("K"))))/(274.15 color(red)(cancel(color(black)("K")))) * "250.9 mL" = "251.8 mL"

The volume of the gas increased by

overbrace("251.8 mL")^(color(blue)("at 275.15 K" = 2^@"C")) - overbrace("250.9 mL")^(color(blue)("at 274.15 K" = 1^@"C")) = "0.9 mL"

You can thus say that with every 1^@"C" = "1 K" increase in temperature, the volume of the gas increases by "0.9 mL".

If you were to draw a graph of this relationship, you'd end up with a straight line that goes "0.9 mL" up as you move "1 K" to the right, i.e. the volume increases by "0.9 mL" with every "1 K" increase in temperature.

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