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Step-by-Step Solution
1. Understanding the Problem
We have 0.1 mole of a monoatomic ideal gas initially at 200 K in a closed vessel. Then 0.05 mole of the same gas at 400 K is added to the vessel. Because the vessel is closed, there is no work done on or by the gas. The process is assumed to be thermally isolated for the mixing, so the total internal energy before mixing equals the total internal energy after mixing.
2. Summarizing Known Data
Initial moles of gas: 0.1 mole at 200\,\text{K}
Moles added: 0.05 mole at 400\,\text{K}
Total final moles of gas: 0.1 + 0.05 = 0.15\,\text{moles}
Gas is monoatomic ideal gas, so C_v = \tfrac{3R}{2}
3. Internal Energy Relation
The internal energy of an ideal monoatomic gas is given by
U = n \times C_v \times T,
and for a monoatomic gas C_v = \tfrac{3R}{2} .
Since no work is done and no heat is exchanged with the surroundings (closed and thermally isolated), the total internal energy before mixing equals the total internal energy after mixing:
U_\text{initial} = U_\text{final}.
4. Computing Internal Energy Before Mixing
The total internal energy before mixing is the sum of the internal energies of both gas samples (though the second sample is being added, we still consider its internal energy at 400 K):
U_\text{initial} =
\Bigl(0.1 \times \tfrac{3R}{2} \times 200 \Bigr)
+ \Bigl(0.05 \times \tfrac{3R}{2} \times 400 \Bigr).
Let us simplify a bit. Factor out \tfrac{3R}{2} if needed, or directly compute in terms of R :
U_\text{initial} =
\tfrac{3R}{2} \times 0.1 \times 200
+ \tfrac{3R}{2} \times 0.05 \times 400.
5. Computing Internal Energy After Mixing
Once equilibrium is reached, the final state has a total of 0.15 moles at the final temperature T_f :
U_\text{final} =
0.15 \times \tfrac{3R}{2} \times T_f.
6. Equating Internal Energies and Solving for T_f
From energy conservation:
U_\text{initial} = U_\text{final}.
Thus,
\Bigl(0.1 \times \tfrac{3R}{2} \times 200 \Bigr)
+ \Bigl(0.05 \times \tfrac{3R}{2} \times 400 \Bigr)
= 0.15 \times \tfrac{3R}{2} \times T_f.
We notice that \tfrac{3R}{2} will cancel out from all terms, so we get:
(0.1 \times 200) + (0.05 \times 400) = 0.15 \, T_f.
This simplifies to:
20 + 20 = 0.15 \, T_f
\quad\Longrightarrow\quad
40 = 0.15 \, T_f.
Hence,
T_f = \frac{40}{0.15} = 266.67 \,\text{K}.
Therefore, the final equilibrium temperature of the gas is approximately 267\,\text{K} .
7. Final Answer
The final equilibrium temperature is about 267\,\text{K} .
