How many moles of #NH_3# are in a 3.0 L vessel at #3*10^2# #K# with a pressure of 1.50 atm?

1 Answer
May 30, 2016

Answer:

There are 0.18 mol of #"NH"_3# in the vessel.

Explanation:

This looks like the time to apply the Ideal Gas Law:

#color(blue)(|bar(ul(PV = nRT)|)#,

where

  • #P# is the pressure
  • #V# is the volume
  • #n# is the number of moles
  • #R# is the gas constant
  • #T# is the temperature

We can rearrange the Ideal Gas Law to get

#n = (PV)/(RT)#

#P = "1.50 atm"#
#V = "3.0 L"#
#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#
#T = 3×10^2 color(white)(l) "K"#

#n = (PV)/(RT) = (1.50 color(red)(cancel(color(black)("atm"))) × 3.0 color(red)(cancel(color(black)("L"))))/( "0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1"))) "mol"^"-1" × 3 × 10^2 color(red)(cancel(color(black)("K")))) = "0.18 mol"#