# Suppose "6g" of "H"_2 and "16 g" of "CH"_4 are in a "1-L" box divided in half by a thin partition, each gas at the same initial pressure P in "atm". Upon mixing the gases, find the total pressure?

May 12, 2017

The question setup leaves a lot of information missing...

But I found the total pressure to be $1.047$ times the initial pressure of either gas, assuming the equipartition theorem applies for both gases and that they are both ideal gases.

Well, as always, a diagram helps...

We are assuming the gases are insulated in their halves of the container before mixing, and the container is itself insulated from the outside.

Since the box is $\text{1 L}$ big, two equal compartments with a thin partition down the middle has, oh, say, $\text{0.5 L}$ in each compartment. Instantaneously removing the partition (by magic?) will mix the two gases, presumably evenly.

By Dalton's law of partial pressures, if we suppose the gases are both ideal:

${P}_{t o t} = {P}_{1} + {P}_{2} + . . .$

If we were to check the $\text{mol}$s of each gas, there were 75% ${\text{H}}_{2}$ and 25% ${\text{CH}}_{4}$. The starting temperatures are (with ${P}_{1 , \text{left") = P_(1,"right}}$:

${T}_{1 , \text{left") = (P_1V_1)/(n_(H_2)R) = (0.5P_1)/("6 g H"_2 xx "1 mol"/("2.016 g H"_2) cdot "0.082057 L"cdot"atm/mol"cdot"K}}$

$= \frac{0.5 {P}_{1}}{\text{2.976 mols H"_2 cdot "0.082057 L"cdot"atm/mol"cdot"K}}$

$= \text{2.047 K/atm} \cdot {P}_{1}$

${T}_{1 , \text{right") = (P_1V_1)/(n_(CH_4)R) = (0.5P_1)/("16 g CH"_4 xx "1 mol"/("16.043 g CH"_4) cdot "0.082057 L"cdot"atm/mol"cdot"K}}$

$= \frac{0.5 {P}_{1}}{\text{0.9973 mols CH"_4 cdot "0.082057 L"cdot"atm/mol"cdot"K}}$

$= \text{6.110 K/atm} \cdot {P}_{1}$

After mixing, ${T}_{2}$ is equal for both gases when they reach thermal equilibrium.

We know that ${V}_{2} = \text{1 L}$ (after the partition is removed). The new partial pressures are based on the new temperature ${T}_{2}$, the same mols, and the new volume.

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

$= \left(\text{2.976 mols H"_2 cdot "0.082057 L"cdot"atm/mol"cdot"K" cdot T_2)/("1 L}\right)$

$= \underline{\text{0.2442 atm/K} \cdot {T}_{2}}$

${P}_{C {H}_{4}} = \frac{{n}_{C {H}_{4}} R {T}_{2}}{V} _ 2$

$= \left(\text{0.9973 mols CH"_4 cdot "0.082057 L"cdot"atm/mol"cdot"K" cdot T_2)/("1 L}\right)$

$= \underline{\text{0.08184 atm/K} \cdot {T}_{2}}$

The total pressure is now written in terms of the thermal equilibrium temperature:

${P}_{2} = {P}_{{H}_{2}} + {P}_{C {H}_{4}}$

$= \underline{\text{0.3260 atm/K} \cdot {T}_{2}}$

At thermal equilibrium, we can find what the new temperature is, based on the gases' heat capacities. For each gas, we assume the equipartition theorem applies, so that:

${C}_{V , {H}_{2}} = \frac{5}{2} R$ $\text{ "" }$ ${C}_{V , C {H}_{4}} = \frac{6}{2} R$

in $\text{J/mol"cdot"K}$.

The heat flow transferred between the two gases is assumed conservative, so:

${q}_{{H}_{2}} + {q}_{C {H}_{4}} = 0$

${n}_{{H}_{2}} {C}_{V , {H}_{2}} \Delta {T}_{{H}_{2}} = - {n}_{C {H}_{4}} {C}_{V , C {H}_{4}} \Delta {T}_{C {H}_{4}}$

$2.976 \cdot \frac{5}{2} R \left({T}_{2} - {T}_{1 , \text{left")) = -0.9973 cdot 6/2R (T_2 - T_(1,"right}}\right)$

We found that

T_(1,"left") = "2.047 K/atm" cdot P_1

T_(1,"right") = "6.110 K/atm" cdot P_1

Therefore,

${T}_{1 , \text{left") = 0.3350T_(1,"right}}$

And so, using $R = \text{8.314 J/mol"cdot"K}$:

$2.976 \cdot \frac{5}{2} R \left({T}_{2} - 0.3350 {T}_{1 , \text{right")) = -0.9973 cdot 6/2R (T_2 - T_(1,"right}}\right)$

$61.8562 \left({T}_{2} - 0.3350 {T}_{1 , \text{right")) = -24.8747(T_2 - T_(1,"right}}\right)$

$61.8562 {T}_{2} - 20.7218 {T}_{1 , \text{right") = -24.8747T_2 + 24.8747T_(1,"right}}$

$\left(61.8562 + 24.8747\right) {T}_{2} = \left(24.8747 + 20.7218\right) {T}_{1 , \text{right}}$

Therefore, the final temperature is:

$\textcolor{b l u e}{{T}_{2}} = \left(\frac{24.8747 + 20.7218}{61.8562 + 24.8747}\right) {T}_{1 , \text{right}}$

$= \textcolor{b l u e}{0.5257 {T}_{1 , \text{right") = 1.569T_(1,"left}}}$

From our expression for ${T}_{1 , \text{right}}$:

${T}_{2} = 0.5257 \left(\text{6.110 K/atm} \cdot {P}_{1}\right)$

$= \text{3.212 K/atm} \cdot {P}_{1}$

As a result, the total pressure becomes:

$\textcolor{b l u e}{{P}_{2}} = \text{0.3260 atm/K} \cdot {T}_{2}$

= "0.3260 atm/K" cdot ("3.212 K/atm"cdot P_1)

$= \textcolor{b l u e}{1.047 {P}_{1}}$

This should make sense, since both gases were treated ideally, so mixing them should keep the total pressure roughly the same as before.

Let's put in some example numbers.

Suppose the starting temperature for ${\text{H}}_{2}$ was $\text{100.00 K}$. It would follow that the temperature for ${\text{CH}}_{4}$ was $\text{298.51 K}$ to be at the same pressure.

This initial pressure is then:

P_1 = T_(1,"left")/("2.047 atm/K") = "48.85 atm"

Physically speaking, methane is indeed still a gas under these conditions, and so is ${\text{H}}_{2}$, so these are realistic choices of temperatures.

The total pressure as a result is found as follows.

${P}_{{H}_{2}} = \text{0.2442 atm/K} \cdot {T}_{2}$

$= \text{0.2442 atm/K"cdot "3.212 K/atm} \cdot {P}_{1}$

$=$ $\text{38.32 atm}$

${P}_{C {H}_{4}} = \text{0.08184 atm/K} \cdot {T}_{2}$

$= \text{0.08184 atm/K"cdot "3.212 K/atm} \cdot {P}_{1}$

$=$ $\text{12.84 atm}$

Therefore, the total pressure is:

${P}_{2} = {P}_{{H}_{2}} + {P}_{C {H}_{4}} = \underline{\text{51.16 atm}}$

The final temperature would then be:

T_2 = 0.5257T_(1,"right") = 0.5257 cdot ("298.51 K") = ul"156.93 K"

This should be between $\text{100.00 K}$ and $\text{298.51 K}$, and it is.

Methane has the higher heat capacity at constant total volume by a factor of $\frac{6 / 2}{5 / 2} = 1.2$, so if the mols were the same, then the final temperature would have lied more towards $\text{298.51 K}$.

However, since there was $2.984$ times as many mols of ${\text{H}}_{2}$, it makes sense that the temperature lies more towards $\text{100.00 K}$.