# What is the temperature, in K, of 2.20 moles of gas occupying 3.5 L at 3.30 atm?

Nov 20, 2016

64 K

#### Explanation:

To figure this out you need to start with the ideal gas equation:

$P \cdot V = n \cdot R \cdot T$

You have P (3.30 atm), V (3.5 l), n (2.20 moles) and you can look up the gas constant, R (0.082057 (latm)/(molK)).

We simply rearrange the ideal gas equation to get T by itself:

$T = \frac{P \cdot V}{n \cdot R}$

Then simply plug in the values you were given, making sure that the units are in liters, atmospheres, and moles (which they are, in this case):

T = (3.30" atm"*3.5" l")/(2.20" moles"*0.082057" (l*atm)/(mol*K)") = 63.9799" K"

Since volume was only given to 2 significant figures, we can only report 2 significant figures for our answer:

64 K.