At #NTP#, for a DIATOMIC gas, how many atoms constitute a #5.6*L# volume?

2 Answers
Jul 15, 2017

Answer:

#2.80xx10^23# #"atoms"#

Explanation:

We're asked to calculate the number of atoms in #5.6# #"L"# of a diatomic gas at normal temperature and pressure (NTP).

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

#PV = nRT#

where

  • #P# is the pressure exerted by the gas, in units of #"atm"# (at NTP, the pressure is defined as #1# #"atm"#

  • #V# is the volume occupied by the gas, in units of #"L"# (given as #5.6# #"L"#)

  • #n# is the number of moles of gas present (we'll need to find this)

  • #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"# (the temperature at NTP is defined as #20^"o""C"#, which is

#20^"o""C" + 273.15 = 293.15# #"K"#)

Plugging in known values, and solving for the quantity, #n#, we have

#n = (PV)/(RT) = ((1cancel("atm"))(5.6cancel("L")))/((0.082057(cancel("L")·cancel("atm"))/("mol"·cancel("K")))(293.15cancel("K"))) = color(red)(0.233# #color(red)("mol gas"#

Using Avogadro's number, we can convert from moles of gas to molecules:

#color(red)(0.233)cancel(color(red)("mol gas"))((6.022xx10^23color(white)(l)"molecules gas")/(1cancel("mol gas")))#

#= 1.40xx10^23# #"molecules gas"#

We're given that the gas is diatomic, meaning there are two atoms per molecule, so the total number of atoms is

#1.40xx10^23cancel("molecules gas")((2color(white)(l)"atoms gas")/(1cancel("molecule gas")))#

#= color(blue)(2.80xx10^23# #color(blue)("atoms"#

Jul 15, 2017

Answer:

Approx. #0.466xxN_A#

Explanation:

So far as I know #"NTP"# specifies conditions of #1*atm#, #293.15*K#, and with this we solve for #n# in the Ideal Gas Equation......

#n=(PV)/(RT)=(1*atmxx5.6*L)/(0.0821*(L*atm)/(K*mol)xx293.15*K)=0.233*mol#.....

But we are not finished there. We were asked for the NUMBER of ATOMS of #X_2# (and in fact most elemental gases are binuclear, certainly the interesting ones), and so we multiply this figure appropriately........

#"Number of atoms"=2xx0.233*molxx6.022xx10^23*mol^-1# #~=1/2N_A#.

Note that looking at last year's exam paper they quoted both #"NTP"# and #"STP"# as supplementary material on the paper.......