What is the volume of the container needed to store 0.8 moles of argon gas at 5.3 atm and 227°C?

2 Answers
Apr 30, 2016

Answer:

#V=(nRT)/P#

Explanation:

#V=(0.8*molxx0.0821*L*atm*K^-1mol^(-1)*500*K)/(5.3*atm)#

#=# #??L#

Apr 30, 2016

Answer:

#=6.19L#

Explanation:

By Ideal gas equation we know,

#PV=nRT#

Where

  • Pressure #P=5.3atm#
  • Volume #V=?L#
  • Universal gas constant #R=0.082LatmK^-1mol^-1#
  • Number moles of gas #n =0.8mol#
  • The temperature of the gas #T=227+273=500K#

Putting inthe gas equation we get

#5.3V=0.8xx0.082xx500#

#V=(0.8xx0.082xx500)/5.3L=6.19L#