(8.8400x10^-1) mol of an ideal gas has a pressure of (1.50x10^2) kPa and temperature of (2.5700x10^2) K. What is the volume of the gas?

May 2, 2017

The volume of the gas is 1.26 × 10^1 color(white)(l)"L".

Explanation:

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

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} p V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

where

• $p$ is the pressure
• $V$ is the volume
• $n$ is the number of moles
• $R$ is the universal gas constant
• $T$ is the temperature

We can rearrange the Ideal Gas Law to get

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

In this problem,

n = 8.8400 × 10^"-1"color(white)(l) "mol"
$R = \text{8.314 kPa·L·K"^"-1""mol"^"-1}$
T = 2.5700 × 10^2 color(white)(l)"K"
p = 1.50 ×10^2 color(white)(l)"kPa"

V = (8.8400 ×10^"-1" color(red)(cancel(color(black)("mol"))) × 8.314 color(red)(cancel(color(black)("kPa")))·"L"·color(red)(cancel(color(black)("K"^"-1"·"mol"^"-1"))) × 2.5700× 10^2color(red)(cancel(color(black)("K"))))/(1.50 × 10^2 color(red)(cancel(color(black)("kPa")))) = 1.26 × 10^1 color(white)(l)"L"