# An ideal gas has a volume of 2.28 L at 279 K and 1.07 atm. What is the pressure when the volume is 1.03 L and the temperature is 307 K?

Dec 29, 2015

$\text{p"= 2.61" atm to 3 significant figures}$

#### Explanation:

First, we use the first set of data to calculate the number of moles using the ideal gas equation:

$\text{pV " = " nRT}$

Where:

• $\text{p}$ is pressure in pascals ($\text{Pa}$)
• $\text{V}$ is volume in cubic metres ( ${\text{m}}^{3}$)
• $\text{n}$ is the number of moles
• $\text{R}$ is the gas constant = $8.314$
• $\text{T}$ is the temperature in Kelvin ($\text{K}$)

First, convert your given values into workable units:

• $1 {\text{L" = 0.001"m"^3, :. 2.28"L" = 0.00228"m}}^{3}$
• $1 \text{atm" = 101325"Pa", :. 1.07"atm" = 108417.8"Pa}$

Second, rearrange the equation to solve for moles:

$\text{n "=" ""pV"/"RT}$

Next, substitute in your given values and calculate the number of moles:

"n "=" "(108417.8"Pa" * 0.00228"m"^3)/(8.314 * 279"K")

$\text{n "=" "0.1065color(red)(666)255" moles}$

We can then move onto calculating the new pressure value. The first thing to do here is to, again, convert non-compliant units into ones that are accepted by the equation:

• $1 {\text{L" = 0.001"m"^3, :. 1.03"L" = 0.00103"m}}^{3}$

Then we rearrange the equation to solve for pressure:

$\text{p "=" ""nRT"/"V}$

And substituting in our values, we get:

"p "=(0.1065666255*8.314 * 307"K")/(0.00103"m"^3)

$\text{p "=264078.0989" Pa" = 2.61" atm to 3 significant figures}$