
The density of methane at $0.20atm$ pressure and $27^\circ C$ is
A: $0.13g{L^{ - 1}}$
B: $0.26g{L^{ - 1}}$
C: $45.67g{L^{ - 1}}$
D: $26.0g{L^{ - 1}}$
Answer
581.1k+ views
Hint: The molecular formula of methane is $C{H_4}$ . We have to consider it to be an ideal gas . To solve this question we will use the ideal gas equation $PV = nRT$ . The density of a given thing can be found out by $\rho = \dfrac{M}{V}$ .
Complete step by step answer:
As we know that ideal gas follows the ideal gas equation. The molecular mass of methane will be $M = 12 + (4 \times 1) = 16$ as the molecular formula of methane is $C{H_4}$ . The ideal gas equation is $PV = nRT$ . In the question the value of pressure and temperature is given. We know the value of molecular mass of methane so
$
\Rightarrow PV = nRT \\
\Rightarrow PV = \dfrac{m}{M}RT \\
\Rightarrow \rho = \dfrac{m}{V} = \dfrac{{PM}}{{RT}} \\
$
Here we have considered that the given mass of the methane is $m$ . So we will now put the known value in the given equation to find the value of the density. We have converted the temperature given to $Kelvin$ scale.
$\rho = \dfrac{{PM}}{{RT}} = \dfrac{{0.20 \times 16}}{{0.0821 \times 300}} = 0.13g{L^{ - 1}}$
So according to the above explanation and calculation the correct answer of the question is option
A: $0.13g{L^{ - 1}}$
Hence option A is correct.
Additional information:
Ideal gases are gases which satisfy the ideal gas equation. The concept of ideal gas is hypothetical.
The molecules of ideal gas follow Newton’s laws of motion. The ideal gas consists of a large number of identical molecules. The collisions between the molecules are considered to be elastic collisions.
The volume of the actual atom of an ideal gas is considered to be zero. The molecules of the ideal gas feel no attraction or repulsion in between them. The molecular speed of the ideal gas is proportional to the Kelvin temperature.
Note:
Whenever we have to find the relation in between temperature , volume , number of molecules , pressure and density always use the ideal gas equation $PV = nRT$ and also remember the density formula $\rho = \dfrac{{PM}}{{RT}}$ .
Complete step by step answer:
As we know that ideal gas follows the ideal gas equation. The molecular mass of methane will be $M = 12 + (4 \times 1) = 16$ as the molecular formula of methane is $C{H_4}$ . The ideal gas equation is $PV = nRT$ . In the question the value of pressure and temperature is given. We know the value of molecular mass of methane so
$
\Rightarrow PV = nRT \\
\Rightarrow PV = \dfrac{m}{M}RT \\
\Rightarrow \rho = \dfrac{m}{V} = \dfrac{{PM}}{{RT}} \\
$
Here we have considered that the given mass of the methane is $m$ . So we will now put the known value in the given equation to find the value of the density. We have converted the temperature given to $Kelvin$ scale.
$\rho = \dfrac{{PM}}{{RT}} = \dfrac{{0.20 \times 16}}{{0.0821 \times 300}} = 0.13g{L^{ - 1}}$
So according to the above explanation and calculation the correct answer of the question is option
A: $0.13g{L^{ - 1}}$
Hence option A is correct.
Additional information:
Ideal gases are gases which satisfy the ideal gas equation. The concept of ideal gas is hypothetical.
The molecules of ideal gas follow Newton’s laws of motion. The ideal gas consists of a large number of identical molecules. The collisions between the molecules are considered to be elastic collisions.
The volume of the actual atom of an ideal gas is considered to be zero. The molecules of the ideal gas feel no attraction or repulsion in between them. The molecular speed of the ideal gas is proportional to the Kelvin temperature.
Note:
Whenever we have to find the relation in between temperature , volume , number of molecules , pressure and density always use the ideal gas equation $PV = nRT$ and also remember the density formula $\rho = \dfrac{{PM}}{{RT}}$ .
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