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Gay Lussac's Law, Avagadro's Law

If you are gi...

Question

If you are given Avogadro's number of atoms of a gas $X$ . If half of the atoms are converted into $X_{(g)}^{+}$ by energy $Δ H$ . The IE of $X$ is :

\frac{2 Δ H}{N_{A}}

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\frac{2 N_{A}}{Δ H}

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\frac{Δ H}{2 N_{A}}

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\frac{N_{A}}{Δ H}

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Solution

The correct option is A $\frac{2 Δ H}{N_{A}}$
Given no. of atoms = Avogadro's no. of atoms = $N_{A} = 6.023 \times 10^{23}$
Given that $\frac{N_{A}}{2}$ atoms are ionized, i.e.,
Ionization energy of $\frac{N_{A}}{2}$ atoms of gas X = $Δ H$
$∴$ Ionization energy of 1 atom of gas X = $\frac{Δ H}{(\frac{N_{A}}{2})} = \frac{2. Δ H}{N_{A}}$
Hence, Ionisation energy of gas X is $\frac{2. Δ H}{N_{A}}$ .

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Q. You are given Avogadro's no. atoms of

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If you are given Avogadro's number of atoms of a gas X. If half of the atoms are converted into X+(g) by energy ΔH. The IE of X is :

The correct option is A 2ΔHNAGiven no. of atoms = Avogadro's no. of atoms = NA=6.023×1023Given that NA2 atoms are ionized, i.e.,Ionization energy of NA2 atoms of gas X = ΔH∴ Ionization energy of 1 atom of gas X = ΔH(NA2)=2.ΔHNAHence, Ionisation energy of gas X is 2.ΔHNA.

If you are given Avogadro's number of atoms of a gas $X$ . If half of the atoms are converted into $X_{(g)}^{+}$ by energy $Δ H$ . The IE of $X$ is :