A container has a volume of 7 L and holds 12 mol of gas. If the container is expanded such that its new volume is 14 L, how many moles of gas must be injected into the container to maintain a constant temperature and pressure?

Jun 30, 2017

$\text{12 mols ideal gas}$ $\underline{\text{added}}$

This is just an ideal gas law problem:

$P V = n R T$

• $P$ is pressure in $\text{atm}$, if $R = \text{0.082057 L"cdot"atm/mol"cdot"K}$.
• $V$ is volume in $\text{L}$.
• $n$ is mols of ideal gas.
• $T$ is temperature in $\text{K}$.

We are told the temperature and pressure must stay constant, and assume a fantastically instantaneous injection of supposedly ideal gas so that the volume doubles. Hence, we can write initial and final states:

$P {V}_{1} = {n}_{1} R T$

$P {V}_{2} = {n}_{2} R T$

or

${V}_{1} / {n}_{1} = {V}_{2} / {n}_{2} = \frac{R T}{P}$

Thus, the final mols of gas are:

${n}_{2} = \left({V}_{2} / {V}_{1}\right) {n}_{1}$

= ("14 L")/("7 L") xx "12 mols"

$=$ $\text{24 mols}$

And we should not be fooled --- the question asked for the CHANGE in the mols of gas, i.e.

$\textcolor{b l u e}{\Delta n} = {n}_{2} - {n}_{1} = \text{24 mols - 12 mols}$

$=$ $\textcolor{b l u e}{\text{12 mols gas injected}}$