# 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?

##### 1 Answer

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

But I found the total pressure to be

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

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

#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, **equal for both gases** when they reach thermal equilibrium.

We know that **new** temperature **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

#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_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

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

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

Methane has the higher heat capacity at constant total volume by a factor of

However, since there was