Question

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

Answer 1

`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"`

Answer 2

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.......