At \[{0^ \circ }C\] the density of nitrogen at 1 atm is 1.25\[\;{\text{kg/}}{{\text{m}}^3}\]. The nitrogen which occupied 1500 ml at \[{0^ \circ }C\] and 1 atm was compressed at \[{0^ \circ }C\] and 575 atm and the gas volume was observed to be 3.92 ml, in violation of Boyle's law. What was the final density of this non-ideal gas?
A) 278 \[\;{\text{kg/}}{{\text{m}}^3}\]
B) 378 \[\;{\text{kg/}}{{\text{m}}^3}\]
C) 478 \[\;{\text{kg/}}{{\text{m}}^3}\]
D) 578 \[\;{\text{kg/}}{{\text{m}}^3}\]
Answer
610.5k+ views
Hint:Density of any substance or the gas is defined as the ratio of mass to volume of the substance or the gas present. It is given by, \[\dfrac{{{\text{mass}}}}{{{\text{volume}}}}.......(1)\]. Unit of mass is Kg and the unit of volume is $m^3/cm^3$.
Complete step by step answer:
We know that from Gas law which is;
$PV = nRT$ Where,
$P$= represents Pressure of gas
$V$= represents Volume of gas
$n$= represents Number of moles
$R$= represents Universal gas constant
$T$= represent the Temperature of gas
According to the given question, let us first find out the number of moles of nitrogen
Therefore from gas law, number of moles of \[{N_2}\] at 273K, 1 atm and 1.5 L and
$R$=${\text{0}}{\text{.0821 L atm }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$
$n = \dfrac{{PV}}{{RT}} = \dfrac{{1 \times 1.5}}{{0.0821 \times 273}} = 0.0669$
Therefore Mass of \[{N_2}\]= number of moles $ \times $molecular mass = 0.0669 $ \times $28 = 1.873g
Given volume at 575 atm
$V$= 3.92 ml = 0.00392 L
To calculate the density which is ratio of mass to volume, values are substituted in equation (1).
Therefore the density of the gas = \[\dfrac{{{\text{mass}}}}{{{\text{volume}}}}{\text{ }} = \dfrac{{1.873}}{{0.00392}} = 477.80\]\[\;{\text{kg/}}{{\text{m}}^3}\]
Therefore, Density of non-ideal gas is = 478.0 \[\;{\text{kg/}}{{\text{m}}^3}\]
Thus the correct option is C.
Note:
Boyle's law, also known as the Boyle–Mariotte law, or another name is Mariotte's law, is defined as an experimental gas law that states how the pressure of a gas tends to change (increase) as the volume of the vessel containing it decreases.
Non-ideal gas: These are the gases which do not follow the perfect gas law which is $PV = nRT$, here p is the pressure, n is the number of moles, R is the gas constant and T is the absolute temperature. And the gases following the perfect gas law are known as perfect or the ideal gases.
Complete step by step answer:
We know that from Gas law which is;
$PV = nRT$ Where,
$P$= represents Pressure of gas
$V$= represents Volume of gas
$n$= represents Number of moles
$R$= represents Universal gas constant
$T$= represent the Temperature of gas
According to the given question, let us first find out the number of moles of nitrogen
Therefore from gas law, number of moles of \[{N_2}\] at 273K, 1 atm and 1.5 L and
$R$=${\text{0}}{\text{.0821 L atm }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$
$n = \dfrac{{PV}}{{RT}} = \dfrac{{1 \times 1.5}}{{0.0821 \times 273}} = 0.0669$
Therefore Mass of \[{N_2}\]= number of moles $ \times $molecular mass = 0.0669 $ \times $28 = 1.873g
Given volume at 575 atm
$V$= 3.92 ml = 0.00392 L
To calculate the density which is ratio of mass to volume, values are substituted in equation (1).
Therefore the density of the gas = \[\dfrac{{{\text{mass}}}}{{{\text{volume}}}}{\text{ }} = \dfrac{{1.873}}{{0.00392}} = 477.80\]\[\;{\text{kg/}}{{\text{m}}^3}\]
Therefore, Density of non-ideal gas is = 478.0 \[\;{\text{kg/}}{{\text{m}}^3}\]
Thus the correct option is C.
Note:
Boyle's law, also known as the Boyle–Mariotte law, or another name is Mariotte's law, is defined as an experimental gas law that states how the pressure of a gas tends to change (increase) as the volume of the vessel containing it decreases.
Non-ideal gas: These are the gases which do not follow the perfect gas law which is $PV = nRT$, here p is the pressure, n is the number of moles, R is the gas constant and T is the absolute temperature. And the gases following the perfect gas law are known as perfect or the ideal gases.
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