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

1 Answer
Nov 20, 2016

Answer:

64 K

Explanation:

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

#P*V = n*R*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 = (P*V)/(n*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.