What volume will 454 grams (1 lb) of hydrogen occupy at 1.05 atm and 25 #"^o#C?

1 Answer
Jul 17, 2017

Answer:

#V = 5.24 xx 10^3# #"L"#

Explanation:

We're asked to find the volume occupied by #454# #"g"# of #"H"_2# at #1.05# #"atm"# and #25# #""^"o""C"#.

To do this, we can use the ideal-gas equation:

#pV = nRT#

where

  • #p# is the pressure, in units of #"atm"# (given as #1.05# #"atm"#)

  • #V# is the volume the gas occupies, in units of #"L"# (we'll be finding this)

  • #n# is the number of moles of the gas present (we'll need to convert the given mass in grams to moles)

  • #R# is the universal gas constant, equal to #0.082057("L"·"atm")/("mol"·"K")#

  • #T# is the absolute temperature of the gas, in units of #"K"# (given #25# #""^"o""C"#)

We need to do some conversions:

Let's find the number of moles of #"H"_2# present via the molar mass of #"H"_2# (#2.02# #"g/mol"#):

#454cancel("g H"_2)((1color(white)(l)"mol H"_2)/(2.02cancel("g H"_2))) = 225# #"mol H"_2#

The temperature in #"K"# is

#25^"o""C" + 273 = 298# #"K"#

Plugging in known values, and solving for the volume, #V#, we have

#V = (nRT)/p = ((225cancel("mol"))(0.082057("L"·cancel("atm"))/(cancel("mol")·cancel("K")))(298cancel("K")))/((1.05cancel("atm")))#

#= color(red)(5.24xx10^3# #color(red)("L"#