Question

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?

Answer 1

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.

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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 `"1 L"` big, two equal compartments with a thin partition down the middle has, oh, say, `"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 ot) = P_1 + P_2 + . . .`

If we were to check the `"mol"`s of each gas, there were `75%` `"H"_2` and `25%` `"CH"_4`. The starting temperatures are (with `P_(1,"left") = P_(1,"right")`:

`T_(1,"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")`

`= (0.5P_1)/("2.976 mols H"_2 cdot "0.082057 L"cdot"atm/mol"cdot"K")`

`= "2.047 K/atm" cdot P_1`

`T_(1,"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")`

`= (0.5P_1)/("0.9973 mols CH"_4 cdot "0.082057 L"cdot"atm/mol"cdot"K")`

`= "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 = "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) = (n_(H_2)RT_2)/V_2`

`= ("2.976 mols H"_2 cdot "0.082057 L"cdot"atm/mol"cdot"K" cdot T_2)/("1 L")`

`= ul("0.2442 atm/K"cdotT_2)`

`P_(CH_4) = (n_(CH_4)RT_2)/V_2`

`= ("0.9973 mols CH"_4 cdot "0.082057 L"cdot"atm/mol"cdot"K" cdot T_2)/("1 L")`

`= ul("0.08184 atm/K"cdotT_2)`

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

`P_2 = P_(H_2) + P_(CH_4)`

`= ul("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) = 5/2R` `" "" "` `C_(V,CH_4) = 6/2R`

in `"J/mol"cdot"K"`.

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

`q_(H_2) + q_(CH_4) = 0`

`n_(H_2)C_(V,H_2)DeltaT_(H_2) = -n_(CH_4)C_(V,CH_4)DeltaT_(CH_4)`

`2.976 cdot 5/2R (T_2 - T_(1,"left")) = -0.9973 cdot 6/2R (T_2 - T_(1,"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,"left") = 0.3350T_(1,"right")`

And so, using `R = "8.314 J/mol"cdot"K"`:

`2.976 cdot 5/2R (T_2 - 0.3350T_(1,"right")) = -0.9973 cdot 6/2R (T_2 - T_(1,"right"))`

`61.8562(T_2 - 0.3350T_(1,"right")) = -24.8747(T_2 - T_(1,"right"))`

`61.8562T_2 - 20.7218T_(1,"right") = -24.8747T_2 + 24.8747T_(1,"right")`

`(61.8562 + 24.8747)T_2 = (24.8747 + 20.7218)T_(1,"right")`

Therefore, the final temperature is:

`color(blue)(T_2) = ((24.8747 + 20.7218)/(61.8562 + 24.8747))T_(1,"right")`

`= color(blue)(0.5257T_(1,"right") = 1.569T_(1,"left"))`

From our expression for `T_(1,"right")`:

`T_2 = 0.5257("6.110 K/atm" cdot P_1)`

`= "3.212 K/atm"cdot P_1`

As a result, the total pressure becomes:

`color(blue)(P_2) = "0.3260 atm/K" cdot T_2`

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

`= color(blue)(1.047P_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 `"H"_2` was `"100.00 K"`. It would follow that the temperature for `"CH"_4` was `"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 `"H"_2`, so these are realistic choices of temperatures.

The total pressure as a result is found as follows.

`P_(H_2) = "0.2442 atm/K"cdotT_2`

`= "0.2442 atm/K"cdot "3.212 K/atm" cdot P_1`

`=` `"38.32 atm"`

`P_(CH_4) = "0.08184 atm/K"cdotT_2`

`= "0.08184 atm/K"cdot "3.212 K/atm" cdot P_1`

`=` `"12.84 atm"`

Therefore, the total pressure is:

`P_2 = P_(H_2) + P_(CH_4) = ul"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 `"100.00 K"` and `"298.51 K"`, and it is.

Methane has the higher heat capacity at constant total volume by a factor of `(6//2)/(5//2) = 1.2`, so if the mols were the same, then the final temperature would have lied more towards `"298.51 K"`.

However, since there was `2.984` times as many mols of `"H"_2`, it makes sense that the temperature lies more towards `"100.00 K"`.