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

Answer:

#"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"#