What is the temperature of 3.3 moles of gas held at a pressure of 12 atm and in a container with a volume of 50 liters?

1 Answer
BRIAN M.
Apr 27, 2018

#2187.9K= T#

Explanation:

This problem requires the Ideal Gas Law equation since we are involving the mole value of the gas.

That equation is
#PV=nRT#
#P=# Pressure in atm
#V=# Volume in L
#n=# moles of gas
#R=0.0831(atm*L)/(mol*K)# - the gas law constant
#T=# Temperature in K

#P=12atm#
#V=50L#
#n=3.3# moles
#R=0.0831(atm*L)/(mol*K)# - the gas law constant
#T=?#

#12atm(50L)=3.3mol(0.0831(atm*L)/(mol*K))T#

#(12atm(50L))/(3.3mol(0.0831(atm*L)/(mol*K)))=T#

#(12cancel(atm)(50cancel(L)))/(3.3cancel(mol)(0.0831(cancel(atm)*cancel(L))/(cancel(mol)*K)))=T#

#(600/(.27423))K= T#

#2187.9K= T#

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