Suppose $6g$ $"6g"$ of $H_{2}$ $"H"_2$ and $16 g$ $"16 g"$ of ${CH}_{4}$ $"CH"_4$ are in a $1-L$ $"1-L"$ box divided in half by a thin partition, each gas at the same initial pressure $P$ $P$ in $atm$ $"atm"$ . Upon mixing the gases, find the total pressure?

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Suppose $6g$ $"6g"$ of $H_{2}$ $"H"_2$ and $16 g$ $"16 g"$ of ${CH}_{4}$ $"CH"_4$ are in a $1-L$ $"1-L"$ box divided in half by a thin partition, each gas at the same initial pressure $P$ $P$ in $atm$ $"atm"$ . Upon mixing the gases, find the total pressure?

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Answer 1 · 2017-05-12T07:57:29

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

But I found the total pressure to be $1.047$ $1.047$ 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 $1 L$ $"1 L"$ big, two equal compartments with a thin partition down the middle has, oh, say, $0.5 L$ $"0.5 L"$ 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:

$P_{t o t} = P_{1} + P_{2} + . . .$ $P_(t ot) = P_1 + P_2 + . . .$

If we were to check the $mol$ $"mol"$ s of each gas, there were $75 %$ $75%$ $H_{2}$ $"H"_2$ and $25 %$ $25%$ ${CH}_{4}$ $"CH"_4$ . The starting temperatures are (with $P_{1, left} = P_{1, right}$ $P_(1,"left") = P_(1,"right")$ :

$T_{1, left} = \frac{P_{1} V_{1}}{n_{H_{2}} R} = \frac{0.5 P_{1}}{{6 g H}_{2} \times \frac{1 mol}{{2.016 g H}_{2}} \cdot 0.082057 L \cdot atm/mol \cdot K}$ $T_(1,"left") = (P_1V_1)/(n_(H_2)R) = (0.5P_1)/("6 g H"_2 xx "1 mol"/("2.016 g H"_2) cdot "0.082057 L"cdot"atm/mol"cdot"K")$

$= \frac{0.5 P_{1}}{{2.976 mols H}_{2} \cdot 0.082057 L \cdot atm/mol \cdot K}$ $= (0.5P_1)/("2.976 mols H"_2 cdot "0.082057 L"cdot"atm/mol"cdot"K")$

$= 2.047 K/atm \cdot P_{1}$ $= "2.047 K/atm" cdot P_1$

$T_{1, right} = \frac{P_{1} V_{1}}{n_{C H_{4}} R} = \frac{0.5 P_{1}}{{16 g CH}_{4} \times \frac{1 mol}{{16.043 g CH}_{4}} \cdot 0.082057 L \cdot atm/mol \cdot K}$ $T_(1,"right") = (P_1V_1)/(n_(CH_4)R) = (0.5P_1)/("16 g CH"_4 xx "1 mol"/("16.043 g CH"_4) cdot "0.082057 L"cdot"atm/mol"cdot"K")$

$= \frac{0.5 P_{1}}{{0.9973 mols CH}_{4} \cdot 0.082057 L \cdot atm/mol \cdot K}$ $= (0.5P_1)/("0.9973 mols CH"_4 cdot "0.082057 L"cdot"atm/mol"cdot"K")$

$= 6.110 K/atm \cdot P_{1}$ $= "6.110 K/atm" cdot P_1$

After mixing, $T_{2}$ $T_2$ is equal for both gases when they reach thermal equilibrium.

We know that $V_{2} = 1 L$ $V_2 = "1 L"$ (after the partition is removed). The new partial pressures are based on the new temperature $T_{2}$ $T_2$ , the same mols, and the new volume.

$P_{H_{2}} = \frac{n_{H_{2}} R T_{2}}{V_{2}}$ $P_(H_2) = (n_(H_2)RT_2)/V_2$

$= \frac{{2.976 mols H}_{2} \cdot 0.082057 L \cdot atm/mol \cdot K \cdot T_{2}}{1 L}$ $= ("2.976 mols H"_2 cdot "0.082057 L"cdot"atm/mol"cdot"K" cdot T_2)/("1 L")$

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

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

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

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

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 $"100.00 K"$ .

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Suppose $6g$ $"6g"$ of $H_{2}$ $"H"_2$ and $16 g$ $"16 g"$ of ${CH}_{4}$ $"CH"_4$ are in a $1-L$ $"1-L"$ box divided in half by a thin partition, each gas at the same initial pressure $P$ $P$ in $atm$ $"atm"$ . Upon mixing the gases, find the total pressure?

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