How many moles of NH_3NH3 are in a 3.0 L vessel at 3*10^23⋅102 KK with a pressure of 1.50 atm?

1 Answer
Ernest Z.
May 30, 2016

There are 0.18 mol of "NH"_3NH3 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"

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