Question

Calculate the density of $ C{{H}_{4}} $ at \[0{}^\circ C\] and 1 atmospheric pressure?

Answer 1

Hint: The density of a pure material is equal to its mass concentration in numerical terms. Varying materials have different densities, which can affect things like buoyancy, purity, and packaging. At normal temperature and pressure, osmium and iridium are the densest known elements. It is frequently substituted with the dimensionless quantity to ease density comparisons across different systems of units.

Complete answer:
Temperature and pressure affect the density of a substance. For solids and liquids, this variance is generally minor, but for gases, it is significantly higher. When you apply more pressure on an object, it shrinks in volume and so becomes denser. With a few exceptions, increasing the temperature of a material reduces its density by increasing its volume. Heating the bottom of a fluid causes heat to convect from the bottom to the top in most materials due to a reduction in the density of the heated fluid. As a result, it rises in comparison to more dense unheated material.
The general gas equation, commonly known as the ideal gas law, is the state equation of a hypothetical ideal gas. Although it has numerous drawbacks, it is a decent approximation of the behaviour of various gases under many situations.
\[PV=nRT\]
P, V, and T denote pressure, volume, and temperature, respectively; n denotes the amount of material; and R denotes the ideal gas constant.
Since $ V=\dfrac{m}{p} $ , where $ m $ is the mass of the gas and $ \rho $ its density, and $ \mathbf{n}=\dfrac{m}{M} $ , where $ M $ is the molar mass of gas, we can write
 $ \dfrac{p_{\mathrm{m}}}{\rho}=\dfrac{m R T}{M} $
We need to solve for $ \rho $ . Dividing both sides by $ P_{\text {m }} $ gives:
 $ \dfrac{1}{\rho}=\dfrac{m R T}{M P_{m}} $
Which then reduces to:
 $ \dfrac{1}{\rho}=\dfrac{R T}{M P^{3}} $
Reciprocating on both sides:
 $ \rho=\dfrac{M P}{R T} $
Here, so
 $ M=16.04 \mathrm{~g} / \mathrm{mol} $
 $ \boldsymbol{p}= $ 1atm
 $ T=273.15 K $
 $ R=0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} $
Inputting we get
 $ \rho=\dfrac{16.04-1}{0.0821-273.15} $
 $ \rho=0.72 $
The units of density are a mass unit divided by a volume unit, as we know. Because the molar mass was stated in grams, the mass unit used here is grams. Because the value of R contained that value, the volume unit is litres.
 $ A=1 \mathrm{~L}=1 \mathrm{dm}^{2} $ , we can write the density as as
 $ \rho =0.72~\text{g}/\text{d}{{\text{m}}^{3}} $ .

Note:
Methane is a chemical molecule with the formula $ C{{H}_{4}} $ that is found in nature (one atom of carbon and four atoms of hydrogen). It is the simplest alkane and a group-14 hydride, and it is the primary component of natural gas. Because of its relative abundance on Earth, methane is a cost-effective fuel, however collecting and storing it is difficult due to its gaseous form under typical temperature and pressure conditions.