Question

A $10.0{\text{ }}{{\text{m}}^3}$ tank is constructed to store LNG (liquefied natural gas, ${\text{CH}}$) at $ - {164^ \circ }{\text{C}}$ and $1{\text{ atm}}$ pressure, under which density is $415{\text{ kg/}}{{\text{m}}^3}$. Calculate the volume of the storage tank capable of holding the same mass of LNG as a gas at ${20^ \circ }{\text{C}}$ and $1{\text{ atm}}$ pressure.

Answer 1

Hint: From the given density first calculate the mass of the LNG. Then calculate the number of moles of LNG. Then using the ideal gas equation calculate the volume of the storage tank.

Formula Used:
1. $d = \dfrac{m}{V}$
2. ${\text{Number of moles}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}}$
3. $PV = nRT$

Complete answer:
First we will calculate the mass of LNG (liquefied natural gas).
We know that the density is the ratio of mass to volume. Thus,
$d = \dfrac{m}{V}$
where
$d$ is the density of the gas,
$m$ is the mass of the gas,
$V$ is the volume of the gas,
Rearrange the equation for the mass of the gas as follows:
$m = d \times V$
Substitute $415{\text{ kg/}}{{\text{m}}^3}$ for the density, $10.0{\text{ }}{{\text{m}}^3}$ for the volume of the gas. Thus,
$m = 415{\text{ kg/}}{{\text{m}}^3} \times 10.0{\text{ }}{{\text{m}}^3}$
$m = 4150{\text{ kg}}$
Thus, the mass of LNG (liquefied natural gas) is $4150{\text{ kg}}$.
Now, calculate the number of moles of LNG (liquefied natural gas) using the equation as follows:
${\text{Number of moles}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}}$
The molar mass of LNG $\left( {{\text{CH}}} \right)$ is the mass of carbon and hydrogen. Thus, molar mass of LNG is $13{\text{ g/mol}}$. Substitute $4150 \times {10^3}{\text{ g}}$ for the mass of LNG. Thus,
${\text{Number of moles}} = \dfrac{{4150 \times {{10}^3}{\text{ g}}}}{{13{\text{ g/mol}}}}$
$\Rightarrow {\text{Number of moles}} = 319.23 \times {10^3}{\text{ mol}}$
Thus, the number of moles of LNG (liquefied natural gas) are $319.23 \times {10^3}{\text{ mol}}$.
We know that the expression for the ideal gas law is as follows:
$PV = nRT$
where,
$P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of moles of gas,
$R$ is the universal gas constant,
$T$ is the temperature of the gas.
Rearrange the equation for the volume of gas as follows:
$V = \dfrac{{nRT}}{P}$
Substitute $319.23 \times {10^3}{\text{ mol}}$ for the number of moles of gas, $0.082{\text{ L atm/K mol}}$ for the universal gas constant, ${20^ \circ }{\text{C}} + 273 = 293{\text{ K}}$ for the temperature, $1{\text{ atm}}$ for the pressure. Thus,
$\Rightarrow V = \dfrac{{319.23 \times {{10}^3}{\text{ mol}} \times 0.082{\text{ L atm/K mol}} \times 293{\text{ K}}}}{{1{\text{ atm}}}}$
$\Rightarrow V = 7.669 \times {10^6}{\text{ L}}$

Thus, the volume of storage tank capable of holding the same mass of LNG as a gas at ${20^ \circ }{\text{C}}$ and $1{\text{ atm}}$ pressure is $7.669 \times {10^6}{\text{ L}} = 7699{\text{ }}{{\text{m}}^3}$.

Note:
We have used the ideal gas law. The ideal gas law states for a given mass of an ideal gas and a constant volume of an ideal gas, the pressure exerted by the molecules of an ideal gas is directly proportional to its absolute temperature.