What is the volume of #2*mol# of an Ideal Gas at #0# #""^@C#, and #1*atm#?

1 Answer
Jul 21, 2017

Answer:

#4.54*10^(-2)m^3#

Explanation:

The ideal gas law says that, #pV=nRT#, where:
#p# = pressure (#Pa#)
#V# = volume (#m^3#)
#n# = number of moles (#mol#)
#R# = gas constant, (#8.31J# #K^(-1)# #mol^(-1)#)
#T# = temperature (#K#)

#1# #atm ~~ 100000Pa#
#0^circC ~~ 273^circK#

#V=(nRT)/P=(2*8.31*273)/100000=0.0453726m^3~~0.0454m^3=4.54*10^(-2)m^3#