# Question #68da5

##### 1 Answer

#### Answer:

#### Explanation:

Your tool of choice here will be the **combined gas law** equation, which looks like this

#color(blue)(|bar(ul(color(white)(a/a)(P_1V_1)/T_1 = (P_2V_2)/T_2color(white)(a/a)|)))" "# , where

Now, before you plug in your values and solve for *relative* to the initial temperature.

Start by assuming that the volume of the gas changes from **pressure**, along with the number of moles of gas, **remain constant**.

In this case, temperature and volume have a **direct relationship** known as **Charles' Law**.

When volume **increases**, temperature **increases** as well, and when volume **decreases**, temperature **decreases** as well.

In your case, the volume *decreases*, so if this was the only change in the conditions of the gas, the new temperature would be **lower** than the initial temperature.

Next, assume that the pressure of the gas changes from **volume** and the number of moles of gas **remain constant**.

When this happens, pressure and temperature have a **direct relationship** known as **Gay Lussac's Law**.

This means that an **increase** in pressure would trigger an **increase** in temperature. If this was the only change in conditions, the new temperature would be **higher** than the initial temperature.

This tells you that *increasing the pressure* of a gas **and** *decreasing its volume* have **competing effects** on its temperature.

Since the decrease in volume is *more significant* than the increase in pressure, you can assume that the final temperature of the gas will be **lower** than the initial temperature.

To see which one "wins", plug in your values into the combined gas law equation. **Do not** forget to convert the initial temperature from *degrees Celsius* to *Kelvin*

#color(purple)(|bar(ul(color(white)(a/a)color(black)(T["K"] = t[""^@"C"] + 273.15)color(white)(a/a)|)))#

Rearrange the equation to solve for

#T_2 = overbrace(P_2/P_1)^(color(blue)("increase in pressure")) * overbrace(V_2/V_1)^(color(red)("decrease in volume")) * T_1#

#T_2 = (770color(red)(cancel(color(black)("torr"))))/(730color(red)(cancel(color(black)("torr")))) * (350color(red)(cancel(color(black)("mL"))))/(450color(red)(cancel(color(black)("mL")))) * (29 + 273.15)"K"#

#T_2 = color(green)(|bar(ul(color(white)(a/a)"250 K"color(white)(a/a)|)))#

If you want, you can express this in *degrees Celsius*

#t_2 = 248 - 273.15 = color(green)(|bar(ul(color(white)(a/a)-23^@"C"color(white)(a/a)|)))#

Both answers are rounded to two **sig figs**.

As predicted, the temperature of the gas **decreased** because the *decrease* in volume was more significant than the *increase* in pressure.