# 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"

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 $\text{1-"""^@"C}$ increase in the temperature of the gas.

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}}}}$

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

${0}^{\circ} \text{C" = 0^@"C" + 273.15 = "273.15 K}$

When the temperature of the gas increases by ${1}^{\circ} \text{C}$, it also increases by $\text{1 K}$ because you have

${0}^{\circ} \text{C" + 1^@"C" = 1^@"C} \to$ the temperatue increases by ${1}^{\circ} \text{C}$

which is

${1}^{\circ} \text{C" = 1^@"C" + 273.15 = "274.15 K}$

Similarly, you will have

$\text{273.15 K" + "1 K" = "274.15 K} \to$ the temperature increases by $\text{1 K}$

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

Rearrange the equation to solve for ${V}_{2}$

${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2} \implies {V}_{2} = {T}_{2} / {T}_{1} \cdot {V}_{1}$

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 $\text{274.15 K}$ to $\text{275.15 K}$. You will once again have--remember to use the volume of the gas at $\text{274.15 mL}$ 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 ${1}^{\circ} \text{C" = "1 K}$ increase in temperature, the volume of the gas increases by $\text{0.9 mL}$.

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