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An insulated container of gas has two champers seperated 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 gas, the final equilibrium temperature of the gas in the container will be
Options
(a) [T₁T₂(P₁V₁+P₂V₂)] / (P₁V₁T₁+P₂V₂T₂)
(b) [T₁T₂(P₁V₁+P₂V₂)] / (P₁V₁T₂+P₂V₂T₁)
(c) (P₁V₁T₁+P₂V₂T₂) / (P₁V₁+P₂V₂)
(d) (P₁V₁T₂+P₂V₂T₁) / (P₁V₁+P₂V₂)
Correct Answer:
[T₁T₂(P₁V₁+P₂V₂)] / (P₁V₁T₂+P₂V₂T₁)
Explanation:
As this is a simple maxing of gas, even if adiabatic conditions are satisfied, PV = nRT for adiabatic as well as isothermal changes. The total number of molecules is conserved.
.·. n₁ = P₁V₁ / R₁T₁, n₂ = P₂V₂ / RT₂
Final state = (n₁ + n₂) RT
(n₁ + n₂) = (P₁V₁ / R₁T₁) + (P₂V₂ / RT₂) = [(T₂P₁V₁+T₁P₂V₂) / RT₁T₂]
T = [(T₁n₁ + T₂n₂) / (n₁ + n₂) = [T₁T₂(P₁V₁+P₂V₂)] / (P₁V₁T₂+P₂V₂T₁)
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Topics: Behavior of Perfect Gas and Kinetic Theory
(34)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The kirchoff’s first law (∑i=0) and second law (∑iR=∑E), where the symbols
- The vector sum of two forces is perpendicular to their vector difference. In that case,
- A wire of length 1 m is moving at a speed of 2 ms⁻¹ perpendicular to its length
- If in a nuclear fusion process, the masses of the fusing nuclei be m₁ and m₂
- If α (current gain) of transistor is 0.98, then what is the value of β
Topics: Behavior of Perfect Gas and Kinetic Theory (34)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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