22.4 L of hydrogen gas contains how many molecules?

2 Answers
Aug 25, 2016

Answer:

At #0°"C"# temperature and #1# atmosphere pressure, this volume is one mole. This is Avogrado's number of molecules, #color(blue)(6.02xx10^{23})#.

Explanation:

The temperature and pressure chosen for this answer, #0°"C"# and #1# atmosphere, were intended to be "standard" temperature and pressure, and the question was apparently formulated with that standard in mind. But the standard pressure was changed to #100# kPa (one atmosphete is #101.3# kPa) in 1982, and confusingly some references still quote the old standard (e.g., http://www.kentchemistry.com/links/GasLaws/STP.htm). One solution to this dilemma that fits with the question is to avoid the "standard" label and refer to the temperature and pressure specifically.

The connection between #22.4" L"# and one mole, at #0°"C"# and #1# atmosphere, comes from the Ideal Gas Law

#PV=nRT#

#P# = pressure, #1# atm
#V# = volume, see below

#n# = moles of gas, put in #1# mole

#R# = gas constant, in pressure-volume unots it's #color(blue)(0.08206 ("L atm")/("mol K"))#

#T# = tempeature, #0°"C" = 273.15 "K"#

So

#(1" atm")(V)=(1" mol")(.08206 ("L atm")/("mol K"))(273.15" K")#

Solve for #V# and to three significant figures you get #color(blue)(22.4" L")#. Thus at #0°"C"# temperature and #1# atmosphere pressure, #22.4" L"# of an ideal gas is one mole.

Aug 26, 2016

Answer:

At a pressure of 1 bar and a temperature of 0 °C, 22.4 L of hydrogen gas contains #5.94 × 10^23color(white)(l) "molecules"#.

Explanation:

The number of molecules depends on the temperature and pressure of the gas.

I shall arbitrarily assume that the gas is at STP (1 bar and 0 °C).

We can use the Ideal Gas Law to calculate the moles of hydrogen:

#color(blue)(|bar(ul(color(white)(a/a)PV = nRT color(white)(a/a)|)))" "#

We can rearrange this to get

#n = (PV)/(RT)#

#P = "1 bar"#
#V = "22.4 L"#
#R = "0.0.08 314 bar·L·K"^"-1""mol"^"-1"#
#T = "(0 + 273.15) K" = "273.15 K"#

#n = (1 color(red)(cancel(color(black)("bar"))) × 22.4 color(red)(cancel(color(black)("L"))))/("0.083 14" color(red)(cancel(color(black)("bar·L·K"^"-1")))"mol"^"-1" × 273.15 color(red)(cancel(color(black)("K")))) = "0.9864 mol"#

We know that 1 mol of a gas contains #6.022 × 10^23 color(white)(l)"molecules"#.

#"No. of molecules" = 0.9864 color(red)(cancel(color(black)("mol"))) × (6.022 × 10^23color(white)(l) "molecules")/(1 color(red)(cancel(color(black)("mol")))) = 5.94 × 10^23color(white)(l) "molecules"#