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

Jul 15, 2017

$2.80 \times {10}^{23}$ $\text{atoms}$

#### Explanation:

We're asked to calculate the number of atoms in $5.6$ $\text{L}$ of a diatomic gas at normal temperature and pressure (NTP).

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

$P V = n R T$

where

• $P$ is the pressure exerted by the gas, in units of $\text{atm}$ (at NTP, the pressure is defined as $1$ $\text{atm}$

• $V$ is the volume occupied by the gas, in units of $\text{L}$ (given as $5.6$ $\text{L}$)

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

• $R$ is the universal gas constant, equal to $0.082057 \left(\text{L"·"atm")/("mol"·"K}\right)$

• $T$ is the absolute temperature of the gas, in units of $\text{K}$ (the temperature at NTP is defined as ${20}^{\text{o""C}}$, which is

${20}^{\text{o""C}} + 273.15 = 293.15$ $\text{K}$)

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")))

$= 1.40 \times {10}^{23}$ $\text{molecules gas}$

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

Approx. $0.466 \times {N}_{A}$

#### Explanation:

So far as I know $\text{NTP}$ specifies conditions of $1 \cdot a t m$, $293.15 \cdot K$, and with this we solve for $n$ in the Ideal Gas Equation......

$n = \frac{P V}{R T} = \frac{1 \cdot a t m \times 5.6 \cdot L}{0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 293.15 \cdot K} = 0.233 \cdot m o l$.....

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........

$\text{Number of atoms} = 2 \times 0.233 \cdot m o l \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$ $\cong \frac{1}{2} {N}_{A}$.

Note that looking at last year's exam paper they quoted both $\text{NTP}$ and $\text{STP}$ as supplementary material on the paper.......