Question

How do you calculate the density of carbon dioxide at $546K$ and $4.00$ atmospheres pressure?

Answer 1

Density is a property of a substance which depends on the other parameters like pressure and temperature. The temperature and pressure change varies the density of gases.

Complete step by step answer:
Let us first understand the density of a substance. Density is defined as the mass of the substance per unit volume. It is denoted by the symbol ρ. The unit of density in the cgs system is $g/mL$ or $g/c{m^3}$ and in the SI system is $Kg/{m^3}$.
Density is mathematically expressed as
$\rho = \dfrac{M}{V}$ where $M$ is the mass of the substance and $V$ is the volume of the substance.
According to the ideal gas equation the pressure and temperature of a gas is related to the volume of the gas. The equation is written as
$PV = nRT$ where $P$ is the pressure of gas, $V$ is the volume of the gas, $n$ is the number of moles of gas, $R$ is the gas constant and $T$is the temperature.
Rearranging the equation,
$V = \dfrac{{nRT}}{P}$.
Thus the density of gas is
$\rho = \dfrac{{PM}}{{RT}}$ (if $n = 1$ for $C{O_2}$ gas)
Given that pressure, $P = 4atm$ and temperature $T = 546K$ . The gas constant $R = 0.0821Latmmo{l^{ - 1}}{K^{ - 1}}$ .
The molar mass of $C{O_2}$ = atomic mass of $C$ + $2 \times$atomic mass of $O$
$= 12 + 2 \times 16 = 44g/mol$
Inserting the values in the equation, the density is
$\rho = \dfrac{{4atm \times 44gmo{l^{ - 1}}}}{{0.0821Latmmo{l^{ - 1}}{K^{ - 1}} \times 546K}} = 3.93g/L$.
Hence the density of carbon dioxide at $546K$ and $4.00$ atmospheres pressure is $3.93g/L$.

Note:
Generally the density of a material changes by the change of pressure or temperature. It is found that increasing the pressure results in increases of the density of a material. Further on increasing the temperature results in a decrease of the density of the material.