What is the pressure of a sample of CH_4 gas (6.022 g) in a 30.0 L vessel at 402 K?

Feb 9, 2017

$\text{106.273524 atm}$

Explanation:

$\text{ Lets recall the gas law}$

$P V = n R T$

$\text{so "P = "(nRT)"/V}$

$\text{Where P = pressure in atm}$
$\text{R = the gas constant 0.0821L}$
$\text{T = temperature in kelvin}$
$\text{n = moles of substance }$
$\text{V = volume in litres}$

Lets now plug in the variables
R = 0.0821L
T = 402K
now for $n$ we have to calculate

$\text{n" = "amount of substance in grams"/"molar mass}$

$\text{n" = "6.022g"/ "16.04 g/mol" = "96.6moles}$

V = 30.0L

$\text{(96.6g * 0.0821 * 402)" / "30.0L} = P r e s s u r e$

$\frac{3188.20572}{\text{30.0L" = "106.273524 atm}}$