Answer
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Hint: In this question, we are asked to calculate the ratio of densities at constant temperature. We are given the ratios of atomic mass and partial pressure of the gas. Now, we know very well that the equation for ideal gas gives us the relation between various parameters such as pressure, volume, amount of substance and temperature. Therefore, we will be using the ideal gas equation to solve this problem.
Formula Used: \[pV=nRT\]
Where
P is the pressure
V is the volume
n is the amount of gas
R is the ideal gas constant
T is the temperature
Complete answer:
From the ideal gas equation
We know,
\[pV=nRT\]
On rearranging the equation
We get,
\[V=\dfrac{nRT}{P}\]
Now, we know that density is given by mass over volume
Therefore, we can say
\[\rho =\dfrac{m}{V}=\dfrac{m}{\dfrac{nRT}{P}}\]
Therefore,
\[\rho =\dfrac{m}{V}=\dfrac{mP}{nRT}\]
We also know that, mass over moles is the molar mass M
Therefore, we can write
\[\rho =\dfrac{MP}{RT}\]
Now, the equation for two gases can be given as
\[{{\rho }_{1}}=\dfrac{{{P}_{1}}{{M}_{1}}}{RT}\]
Similarly, for second gas
\[{{\rho }_{2}}=\dfrac{{{P}_{2}}{{M}_{2}}}{RT}\]
Now, it is given that temperature is constant. Also, R is always constant
Therefore, on taking the ratio of density of two gases
From (1) and (2) we get,
\[\dfrac{{{\rho }_{1}}}{{{\rho }_{2}}}=\dfrac{{{P}_{1}}{{M}_{1}}}{{{P}_{2}}{{M}_{2}}}\]
After substituting value
We get,
\[\dfrac{{{\rho }_{1}}}{{{\rho }_{2}}}=\dfrac{2\times 4}{3\times 3}\]
On solving
\[\dfrac{{{\rho }_{1}}}{{{\rho }_{2}}}=\dfrac{8}{9}\]
Therefore, the correct answer is option D.
Note:
A monatomic gas is the gas having only one atom and lacks any covalent bond. All the noble gases are said to be monatomic. The gases that obey the ideal gas law are said to be the ideal gases. These are hypothetical gases which are assumed to behave ideally like a gas. Ideal gas law is an equation which states that pressure, volume and mass of the gases are related to each other. The ideal gas law has many limitations as gases real gases do not behave ideally. The variations of Gas laws are the Charles law, Boyle's Law, Gauss’s law.
Formula Used: \[pV=nRT\]
Where
P is the pressure
V is the volume
n is the amount of gas
R is the ideal gas constant
T is the temperature
Complete answer:
From the ideal gas equation
We know,
\[pV=nRT\]
On rearranging the equation
We get,
\[V=\dfrac{nRT}{P}\]
Now, we know that density is given by mass over volume
Therefore, we can say
\[\rho =\dfrac{m}{V}=\dfrac{m}{\dfrac{nRT}{P}}\]
Therefore,
\[\rho =\dfrac{m}{V}=\dfrac{mP}{nRT}\]
We also know that, mass over moles is the molar mass M
Therefore, we can write
\[\rho =\dfrac{MP}{RT}\]
Now, the equation for two gases can be given as
\[{{\rho }_{1}}=\dfrac{{{P}_{1}}{{M}_{1}}}{RT}\]
Similarly, for second gas
\[{{\rho }_{2}}=\dfrac{{{P}_{2}}{{M}_{2}}}{RT}\]
Now, it is given that temperature is constant. Also, R is always constant
Therefore, on taking the ratio of density of two gases
From (1) and (2) we get,
\[\dfrac{{{\rho }_{1}}}{{{\rho }_{2}}}=\dfrac{{{P}_{1}}{{M}_{1}}}{{{P}_{2}}{{M}_{2}}}\]
After substituting value
We get,
\[\dfrac{{{\rho }_{1}}}{{{\rho }_{2}}}=\dfrac{2\times 4}{3\times 3}\]
On solving
\[\dfrac{{{\rho }_{1}}}{{{\rho }_{2}}}=\dfrac{8}{9}\]
Therefore, the correct answer is option D.
Note:
A monatomic gas is the gas having only one atom and lacks any covalent bond. All the noble gases are said to be monatomic. The gases that obey the ideal gas law are said to be the ideal gases. These are hypothetical gases which are assumed to behave ideally like a gas. Ideal gas law is an equation which states that pressure, volume and mass of the gases are related to each other. The ideal gas law has many limitations as gases real gases do not behave ideally. The variations of Gas laws are the Charles law, Boyle's Law, Gauss’s law.
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