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

1 Answer
Feb 9, 2017

"106.273524 atm"

Explanation:

" Lets recall the gas law"

PV = nRT

"so "P = "(nRT)"/V"

"Where P = pressure in atm"
"R = the gas constant 0.0821L"
"T = temperature in kelvin"
"n = moles of substance "
"V = volume in litres"

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

"n" = "amount of substance in grams"/"molar mass"

"n" = "6.022g"/ "16.04 g/mol" = "96.6moles"

V = 30.0L

"(96.6g * 0.0821 * 402)" / "30.0L" = Pressure

3188.20572 / "30.0L" = "106.273524 atm"