The Question is from neet-physics thermodynamics , and it was created by the author VENKATESWARARAO

Question

Two identical isolated containers A and B contain ideal gases with the same number of moles. Container A is at a temperature $T_A$ while container B is at a temperature $T_B = 2T_A$.  A third identical container C is evacuated and brought into thermal contact with A and then with B.  If the final temperature of C is $T_C$, which of the following relations is correct?

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
	**Primary Concept**: Heat transfer and thermal equilibrium.
	**Related Topics**: Ideal gas law, specific heat capacity.
	**Prerequisites**: Understanding of thermal equilibrium, heat transfer, and how temperature changes when objects are brought into contact.
## SOLUTION PATHWAY
**Given Information**:
	* Container A: Temperature $T_A$, number of moles $n_A = n$
	* Container B: Temperature $T_B = 2T_A$, number of moles $n_B = n$
	* Container C: Initially evacuated, $n_C = 0$, identical to A and B.
	* Identical containers imply equal heat capacities at constant volume.
**Method & Steps**:
	1. **Thermal contact between A and C**:
		Reasoning: When two objects are brought into thermal contact, heat flows from the hotter object to the colder object until they reach thermal equilibrium. Heat lost by A will be gained by C.
		Formula/Rule: $Q=nC_v\Delta T$, where $Q$ is heat, $n$ is number of moles, $C_v$ is molar heat capacity at constant volume, and $\Delta T$ is change in temperature.
	2. **Equilibrium Temperature between A and C**:
		Working: Let $T_{AC}$ be the equilibrium temperature between A and C. Heat lost by A = Heat gained by C. So, $nC_v(T_A - T_{AC}) = nC_v(T_{AC} - 0)$. This simplifies to $T_A - T_{AC} = T_{AC}$, thus $2T_{AC} = T_A$ and $T_{AC} = T_A/2$.  Now, container C has temperature $T_A/2$.
		Key Point: The number of moles in C is equal to the number of moles in A and B after the contact. Initially, C is evacuated, but after contact with A, it will have the same number of moles, n. So, the heat gained by C is for the same amount of matter as A.
	3. **Thermal contact between C and B**:
		Reasoning: Now, container C with temperature $T_A/2$ is brought into thermal contact with B with temperature $2T_A$. Heat flows from B to C.
	4. **Final Equilibrium Temperature**:
		Working: Let $T_C$ be the final equilibrium temperature. Heat lost by B = Heat gained by C. So, $nC_v(2T_A - T_C) = nC_v(T_C - T_A/2)$. This simplifies to $2T_A - T_C = T_C - T_A/2$. Thus, $2T_C = 2T_A + T_A/2 = (5/2)T_A$, and finally $T_C = (5/4)T_A$.
		Result: The final temperature of C is $T_C = (5/4)T_A$.
		Verification: The final temperature should be between $T_A/2$ and $2T_A$, and $(5/4)T_A$ satisfies this condition.
 ## OPTION ANALYSIS
[A] :x: The final temperature is not equal to $T_A$. The temperature of C first becomes $T_A/2$ after contact with A, and then rises further after contact with B.
[B] :x: The final temperature is not equal to $(3/2)T_A$.
[C] ✓: This is the correct final temperature of container C, calculated in the solution path.
[D] :x: The final temperature is not equal to $2T_A$.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Assuming that the final temperature is the average of the initial temperatures. This is incorrect because the heat capacities are equal, but the initial temperatures are different and there are two steps in the process.
		2.[Error 2]: Forgetting that container C is initially evacuated and then acquires moles of gas when it comes into contact with A.
	**Quick Tips**:
		1.[Helpful shortcut]: In thermal equilibrium problems, remember that the heat lost by the hot object is equal to the heat gained by the cold object.
		2.[Memory aid]:  Always think step by step, considering the heat transfer during each contact.

VENKATESWARARAO · Answer

T_C = T_A

VENKATESWARARAO · Answer

T_C = (3/2)T_A

VENKATESWARARAO · Answer

The correct Option is "T_C = (5/4)T_A"

VENKATESWARARAO · Answer

T_C = 2T_A

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