The Question is from neet-physics kinetic-theory-gases , and it was created by the author ABDUL JILANI BASHA

Question

A mixture of two ideal diatomic gases A and B, initially at temperatures $T_A$ and $T_B$ respectively, is allowed to reach thermal equilibrium at constant volume.  If $n_A$ and $n_B$ are the number of moles of A and B, what is the final equilibrium temperature?

ABDUL JILANI BASHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: [C]
## CONCEPT FRAMEWORK
	**Primary Concept**: Thermal Equilibrium and Energy Conservation
	**Related Topics**:  Heat transfer, Specific Heat Capacity, Internal Energy of Ideal Gases
	**Prerequisites**: Understanding of ideal gas behavior, equipartition theorem, heat transfer, and the concept of thermal equilibrium.
## SOLUTION PATHWAY
**Given Information**:
	* Two ideal diatomic gases A and B.
	* Initial temperatures: $T_A$ and $T_B$.
	* Number of moles: $n_A$ and $n_B$.
	* Constant volume process.
	* Reaching thermal equilibrium (final temperature $T_f$)
**Method & Steps**:
	1. **Internal Energy of Diatomic Gas:**
		Reasoning: The internal energy of a diatomic ideal gas is given by $U = \frac{5}{2} n R T$, where n is number of moles, R is the gas constant and T is temperature.
		Formula/Rule: $U = \frac{5}{2} n R T$
	2. **Initial Total Internal Energy:**
		Working: The initial total internal energy is the sum of the internal energies of gases A and B.
		Key Point: $U_{initial} = U_A + U_B = \frac{5}{2} n_A R T_A + \frac{5}{2} n_B R T_B$
	3. **Final Total Internal Energy:**
		Working: After reaching equilibrium at $T_f$, the final total internal energy is:
		Key Point: $U_{final} = \frac{5}{2} (n_A + n_B) R T_f$
	4. **Energy Conservation:**
		Working: Since the volume is constant and no work is done, the total internal energy is conserved.
		Key Point: $U_{initial} = U_{final}$
	5. **Solve for Final Temperature:**
		Working: $\frac{5}{2} n_A R T_A + \frac{5}{2} n_B R T_B = \frac{5}{2} (n_A + n_B) R T_f$
		Working: Canceling out $\frac{5}{2} R$ gives: $n_A T_A + n_B T_B = (n_A + n_B) T_f$
		Result: $T_f = \frac{n_A T_A + n_B T_B}{n_A + n_B}$
		Verification: The final temperature is the weighted average of the initial temperatures.
 ## OPTION ANALYSIS
[A] :x: This is incorrect as it represents a geometric mean, not the correct weighted average of temperatures.
[B] :x: This is incorrect as it calculates the simple average of temperatures, ignoring the number of moles.
[C] ✓: This is the correct formula derived from energy conservation and the definition of internal energy.
[D] :x: This has an additional factor of 2, which is incorrect.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Using simple average of temperatures instead of weighted average. 
			Solution: Remember the number of moles influence the final temperature.
		2.[Error 2]: Forgetting the form of internal energy of diatomic gases.
			Solution: Remember to use $U = \frac{5}{2}nRT$ for diatomic gases
	**Quick Tips**:
		1.[Helpful shortcut]: In problems with thermal equilibrium, always start by equating initial and final total energies.
		2.[Memory aid]: Remember the internal energy for diatomic gases is proportional to $5/2$ times the number of moles and temperature.

ABDUL JILANI BASHA · Answer

$√(T_A T_B)$

ABDUL JILANI BASHA · Answer

$(T_A + T_B)/2$

ABDUL JILANI BASHA · Answer

The correct Option is "$(n_A T_A + n_B T_B) / (n_A + n_B)$"

ABDUL JILANI BASHA · Answer

$2(n_A T_A + n_B T_B) / (n_A + n_B)$

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