
The density of a gaseous substance at \[1{\text{ atm}}\] and $750{\text{ K}}$ is $0.30{\text{ g/lt}}$. It the molecular weight of the substance is ${\text{27}}$, the dominant forces existing among gas molecule is-
A.Attractive
B.Repulsive
C.Both (A) and (C)
D.None of these
Answer
578.1k+ views
Hint:The dominant forces existing among the gases are related to the compressibility factor. The factor that describes deviation of a gas from the ideal gas behaviour is known as the compressibility factor.
Complete step by step answer:
Calculate the compressibility factor using the equation as follows:
$PV = ZnRT$ …… (1)
Where $P$ is the pressure of the gas,
$V$ is the volume of the gas,
$Z$ is the compressibility factor,
$n$ is the number of moles of gas,
$R$ is the universal gas constant,
$T$ is the temperature.
We know that,
$d = \dfrac{m}{V}$
Where $d$ is the density of the gas,
$m$ is the molecular weight of the gas,
$V$ is the volume of the gas.
Thus,
$V = \dfrac{m}{d}$
Thus, equation (1) becomes as follows:
$P \times \dfrac{m}{d} = ZnRT$
$Pm = ZdnRT$
$Z = \dfrac{{Pm}}{{dnRT}}$
Substitute \[1{\text{ atm}}\] for the pressure, ${\text{27 g/mol}}$ for the molecular weight of the gas, $0.30{\text{ g/lt}}$ for the density of the gas, $1{\text{ mol}}$ for the number of moles of gas, $0.082{\text{ L atm/mol K}}$ for the universal gas constant, $750{\text{ K}}$ for the temperature. Thus,
$Z = \dfrac{{1{\text{ atm}} \times {\text{27 g/mol}}}}{{0.30{\text{ g/lt}} \times 1{\text{ mol}} \times 0.082{\text{ L atm/mol K}} \times 750{\text{ K}}}}$
$Z = 1.46$
Thus, the compressibility factor is ${\text{1}}{\text{.46}}$.
Here, the compressibility factor is greater than one. Thus, the dominant forces existing among gas molecules are repulsive.
Thus the correct option is option (B).
Note:
When the compressibility factor is greater than one, the gas is more compressible and thus, the forces existing are repulsive. When the compressibility factor is less than one, the gas is more expandable and thus, the forces existing are attractive. For an ideal gas, compressibility factor is always equal to one.
Complete step by step answer:
Calculate the compressibility factor using the equation as follows:
$PV = ZnRT$ …… (1)
Where $P$ is the pressure of the gas,
$V$ is the volume of the gas,
$Z$ is the compressibility factor,
$n$ is the number of moles of gas,
$R$ is the universal gas constant,
$T$ is the temperature.
We know that,
$d = \dfrac{m}{V}$
Where $d$ is the density of the gas,
$m$ is the molecular weight of the gas,
$V$ is the volume of the gas.
Thus,
$V = \dfrac{m}{d}$
Thus, equation (1) becomes as follows:
$P \times \dfrac{m}{d} = ZnRT$
$Pm = ZdnRT$
$Z = \dfrac{{Pm}}{{dnRT}}$
Substitute \[1{\text{ atm}}\] for the pressure, ${\text{27 g/mol}}$ for the molecular weight of the gas, $0.30{\text{ g/lt}}$ for the density of the gas, $1{\text{ mol}}$ for the number of moles of gas, $0.082{\text{ L atm/mol K}}$ for the universal gas constant, $750{\text{ K}}$ for the temperature. Thus,
$Z = \dfrac{{1{\text{ atm}} \times {\text{27 g/mol}}}}{{0.30{\text{ g/lt}} \times 1{\text{ mol}} \times 0.082{\text{ L atm/mol K}} \times 750{\text{ K}}}}$
$Z = 1.46$
Thus, the compressibility factor is ${\text{1}}{\text{.46}}$.
Here, the compressibility factor is greater than one. Thus, the dominant forces existing among gas molecules are repulsive.
Thus the correct option is option (B).
Note:
When the compressibility factor is greater than one, the gas is more compressible and thus, the forces existing are repulsive. When the compressibility factor is less than one, the gas is more expandable and thus, the forces existing are attractive. For an ideal gas, compressibility factor is always equal to one.
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