# Question #c046e

##### 1 Answer

#### Answer:

#### Explanation:

Before doing any calculation, try to predict what you the new pressure of the gas will be *relative* to its initial value.

The problem tells you that

A sample of argon... is cooledin the same containerto a...

This tells you two important things

the volume of the gas remains unchangedthe number of moles of gas remains constant

When number of moles and volume are kept constant, *pressure* and *temperature* have a **direct relationship** - this is known as Gay Lussac's Law.

Simply put, when temperature **decreases**, pressure **decreases** as well, and when temperature **increases**, pressure **increases** as well.

In your case, the gas is *cooled*, so you can definitely expect a **decrease** in its pressure.

Mathematically, this is written like this

#color(blue)(P_1/T_1 = P_2/T_2)" "# , where

Now, it is important to remember that the temperature of the gas **must be** expressed in *Kelvin*!

Plug in your values into the above equation and solve for

#P_1/T_1 = P_2/T_2 implies P_2 = T_2/T_1 * P_1#

#P_2 = ( (273.15 + 0) color(red)(cancel(color(black)("K"))))/( (273.15 + 300) color(red)(cancel(color(black)("K")))) * "50.0 atm"#

#P_2 = "23.83 atm"#

Now, this *should* be rounded off to one sig fig, the number of sig figs you have for the final temperature of the gas, but I'll leave it rounded to two sig figs

#P_2 = color(green)("24 atm")#