A closed vessel contains 0.1 mole of a monoatomic ideal gas at 200 K. If 0.05 mole of the same gas at 400 K is added to it, the final equilibrium temperature (in K) of the gas in the vessel will be close to _______. (Round off to the nearest integer)

Solution

Physics sir · 4 steps

No heat escapes and no work is done, so the total internal energy of the two lots of gas stays fixed.

1. Energy is conserved. The vessel is closed and rigid, so the gas does no work and exchanges no heat with the outside. The total internal energy before mixing equals the total after mixing.
2. Internal energy. For an ideal gas U = nC_VT, with C_V = 3/2R for a monoatomic gas. Both portions are the same gas, so C_V is the same on both sides and cancels.

   n₁T₁ + n₂T₂ = (n₁ + n₂) T

3. Put in numbers. 0.1 × 200 + 0.05 × 400 = 20 + 20 = 40. Divide by the total moles: T = 40/0.15 = 266.7 K.
4. So. The final temperature is about 266.7 K, which rounds to 267.

The common slip: 300. Taking the plain average of 200 K and 400 K gives 300, but that ignores that the 200 K gas has twice as many moles.

The idea to keep. When two samples of the same ideal gas mix with no heat loss and no work, the final temperature is the mole-weighted average of the starting temperatures.

Answer from the NTA final answer key. Solution by PrepWiser.