# 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?

May 30, 2016

#### Answer:

There are 0.18 mol of ${\text{NH}}_{3}$ 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

$n = \frac{P V}{R T}$

$P = \text{1.50 atm}$
$V = \text{3.0 L}$
$R = \text{0.082 06 L·atm·K"^"-1""mol"^"-1}$
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"