Question

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

Answer 1

Hint: 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.