At $N T P$ $NTP$ , for a DIATOMIC gas, how many atoms constitute a $5.6 \cdot L$ $5.6*L$ volume?

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At $N T P$ $NTP$ , for a DIATOMIC gas, how many atoms constitute a $5.6 \cdot L$ $5.6*L$ volume?

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Answer 1 · 2017-07-15T04:45:07

$2.80 \times 10^{23}$ $2.80xx10^23$ $atoms$ $"atoms"$

Explanation:

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

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

$P V = n R T$ $PV = nRT$

where

$P$ $P$ is the pressure exerted by the gas, in units of $atm$ $"atm"$ (at NTP, the pressure is defined as $1$ $1$ $atm$ $"atm"$
$V$ $V$ is the volume occupied by the gas, in units of $L$ $"L"$ (given as $5.6$ $5.6$ $L$ $"L"$ )
$n$ $n$ is the number of moles of gas present (we'll need to find this)
$R$ $R$ is the universal gas constant, equal to $0.082057 \frac{L \cdot atm}{mol \cdot K}$ $0.082057("L"·"atm")/("mol"·"K")$
$T$ $T$ is the absolute temperature of the gas, in units of $K$ $"K"$ (the temperature at NTP is defined as $20^{o} C$ $20^"o""C"$ , which is

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

Plugging in known values, and solving for the quantity, $n$ $n$ , 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 · 2017-07-15T04:55:49

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.......

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At $N T P$ $NTP$ , for a DIATOMIC gas, how many atoms constitute a $5.6 \cdot L$ $5.6*L$ volume?

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Explanation:

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At NTPNTPNTP, for a DIATOMIC gas, how many atoms constitute a 5.6*L5.6⋅L5.6*L volume?

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Explanation:

Explanation:

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At $N T P$ $NTP$ , for a DIATOMIC gas, how many atoms constitute a $5.6 \cdot L$ $5.6*L$ volume?