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?

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"# 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.