Question

How many litres would 14 grams of chlorine gas occupy at $300.0{\text{K}}$ and $1.51 {\text{atm}}$ ?

Answer 1

Using the molar mass of chlorine gas to determine how many moles of chlorine gas you have in that sample is the first thing to do here. You can use the ideal gas law equation to solve the volume of the sample now that you know how many moles of gas you are dealing with, and the conditions of pressure and temperature at which the gas is held.

Formula Used:
We will use the formula for ideal gas law to solve this question
$PV = nRT$
Where
$P$ is the pressure of the gas
$V$ is the volume occupied by the gas
$n$ is the number of moles of the gas
$R$ is the gas constant usually given as $0.0821\dfrac{{{\text{ atm}} \cdot {\text{L }}}}{{{\text{mol}} \cdot {\text{K}}}}$
$T$ is the temperature of the gas, considered in kelvin.

Complete Step-by-Step Solution
According to the question, the following information is provided to us
The temperature of the gas, $T = 300 {\text{K}}$
The pressure of the gas, $P = 1.51 {\text{atm}}$
The formula for chlorine gas is $C{l_2}$
The molecular weight of chlorine gas, $C{l_2}$ is $70 {\text{g mo}}{{\text{l}}^{{\text{ - 1}}}}$
Now, let us calculate the number of moles of chlorine gas in the sample
This can be calculated by the following expression
Since $70 {\text{g}}$ contains $1 {\text{mole}} {\text{C}}{{\text{l}}_2}$
Then, 14g would contain,
$\dfrac{{1{\text{mole}} {\text{C}}{{\text{l}}_2}}}{{70 {\text{g}}}} \times 14 {\text{g}} = 0.2 {\text{moles C}}{{\text{l}}_2}$
Since, we know the number of moles of $C{l_2}$
We can now easily use the formula above to find the volume of the sample
$PV = nRT$
Rewriting the equation will give us the formula for volume
$V = \dfrac{{nRT}}{P}$
Let us now substitute the values into the above expression
$V = \dfrac{{0.2{\text{ moles}} \times 0.0821\dfrac{{{\text{ atm}} \cdot {\text{L}}}}{{{\text{ mol}} \cdot {\text{K}}}} \times 300.0{\text{K}}}}{{1.51 {\text{atm}}}} = 3.262{\text{L}}$
Hence, the required volume of chlorine gas at $300.0{\text{K}}$ and $1.51 {\text{atm}}$ is $3.262{\text{L}}$ .

Note:
The ideal gases that have elastic collisions between their molecules are ideal gases and there are no attractive intermolecular forces. There is no such thing as ideal gases in reality. Under certain conditions of temperature and pressure, the gases just show ideal behaviour.