Question

An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume V₁ and contains ideal gas at pressure P₁ and temperature T₁. The other chamber has volume V₂ and contains ideal gas at pressure P₂ and temperature T₂. If the partition is removed without doing any work on the gases, calculate the final equilibrium temperature of the container.

Answer 1

Let T be the equilibrium temperature and let n₁ and n₂ be the number of moles in vessels 1 and 2 respectively. As there is no loss of energy,

`"n"_1(3/2"RT"_1) + "n"_2(3/2"RT"_2) = ("n"_1 + "n"_2) (3/2"RT")`

`"n"_1"T"_1 + "n"_2"T"_2 = ("n"_1 + "n"_2)"T"`

T = `("n"_1"T"_1 + "n"_2"T"_2)/("n"_1 + "n"_2)`

Now, n₁ = `("P"_1"V"_1)/"RT"_1`, n₂ = `("P"_2"V"_2)/"RT"_2`

Substituting n₁ and n₂ values and solving, we get

T = `("T"_1"T"_2("P"_1"V"_1 + "P"_2"V"_2))/("P"_1"V"_1"T"_2 + "P"_2"V"_2"T"_1)`