# If the volume of the container is "1.00 L", what is the change in mols of gas required to change the pressure from "1.00 atm" to "0.920 atm" if the temperature was also dropped from 25^@ "C" to 22^@ "C"? (R = "0.082057 L"cdot"atm/mol"cdot"K")

Jun 1, 2016

I get about $\Delta n = {n}_{2} - {n}_{1} = \text{0.00289 mols}$.

Since you are given:

• $V = \text{1.00 L}$, the volume
• ${P}_{1} = \text{1.00 atm}$, the initial pressure
• ${P}_{2} = \text{0.920 atm}$, the final pressure
• ${T}_{1} = {25}^{\circ} \text{C}$, the initial temperature
• ${T}_{2} = {22}^{\circ} \text{C}$, the final temperature
• $R = \text{0.082057 L"cdot"atm/mol"cdot"K}$, the universal gas constant

we know that the volume had to have remained constant (we only got one value for it).

Also, since we were given pressure, temperature, volume, and $R$, it's a good hint that we are using the ideal gas law:

$P V = n R T$

Since we have the initial and final values for pressure and temperature, we have two variations on the ideal gas law:

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

So finding "the mols" should mean finding the change in the mols, ${n}_{2} - {n}_{1} = \Delta n$.

A reasonable way to do this is to solve for the initial $\text{mol}$s:

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

$= \left(\left(\text{1.00 atm")("1.00 L"))/(("0.082057 L"cdot"atm/mol"cdot"K")("25 + 273.15 K}\right)\right)$

$=$ $\text{0.0409 mols}$

So, ${n}_{2}$ is gotten as follows:

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

$= \left(\left(\text{0.920 atm")("1.00 L"))/(("0.082057 L"cdot"atm/mol"cdot"K")("22 + 273.15 K}\right)\right)$

$=$ $\text{0.0380 mols}$

So, if you wanted the change in $\text{mol}$s, you would get:

$\textcolor{b l u e}{\Delta n}$

$=$ $\text{0.0409 - 0.0380 mols}$

$=$ $\textcolor{b l u e}{\text{0.00289 mols}}$