What is the volume of 1.00 * 10^241.001024 gg of CH_4CH4 gas at STPSTP?

1 Answer
May 27, 2016

The volume is 1.42 × 10^15color(white)(l) "km"^31.42×1015lkm3.

We can convert the mass to moles of methane and then use the Ideal Gas Law to calculate the volume at STP.

Moles of "CH"_4CH4

"Moles of CH"_4 = 1.00 × 10^24 color(red)(cancel(color(black)("g CH"_4))) × ("1 mol CH"_4)/(16.04 color(red)(cancel(color(black)("g CH"_4)))) = 6.234 × 10^22 color(white)(l)"mol CH"_4

Volume at STP

The Ideal Gas Law is:

color(blue)(|bar(ul(PV = nRT)|),

where

  • P is the pressure
  • V is the volume
  • n is the number of moles
  • R is the gas constant
  • T is the temperature

We can rearrange the Ideal Gas Law to get

V = (nRT)/P

STP is 1 bar and 0 °C.

n = 6.234 × 10^22color(white)(l) "mol"
R = "0.083 14 bar·L·K"^"-1""mol"^"-1"
T = "273.15 K"
P = "1 bar"

V = (nRT)/P = (6.234 × 10^22 color(red)(cancel(color(black)("mol"))) × "0.083 14" color(red)(cancel(color(black)("bar")))·"L"· color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 273.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("bar")))) = 1.42 × 10^24 color(white)(l)"L" = 1.42 × 10^21color(white)(l) "m"^3 = 1.42 × 10^15color(white)(l) "km"^3

This is almost exactly the volume of Jupiter.