The Question is from neet-physics thermodynamics internal-energy-1st-law-of-thermodynamics, and it was created by the author VENKATESWARARAO

Question

A rigid, insulated container is divided into two compartments by a partition. One compartment contains 2 moles of an ideal monatomic gas at temperature $T_1$, and the other contains 1 mole of a diatomic ideal gas at temperature $T_2$. If the partition is removed, and the gases are allowed to mix and reach equilibrium, what is the final temperature of the mixture? Assume no heat exchange with the surroundings.

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Conservation of Energy
	**Related Topics**: Thermodynamics, Ideal Gas Law, Internal Energy of Ideal Gases
	**Prerequisites**: Understanding of heat transfer, specific heat capacities of ideal gases, internal energy change.
## SOLUTION PATHWAY
**Given Information**:
	* Number of moles of monatomic gas: $n_1 = 2$
	* Initial temperature of monatomic gas: $T_1$
	* Number of moles of diatomic gas: $n_2 = 1$
	* Initial temperature of diatomic gas: $T_2$
	* The system is insulated, meaning no heat exchange with surroundings.
	* The container is rigid, so no work is done (constant volume)
**Method & Steps**:
	1.	Determine the initial internal energy of each gas.
		Reasoning: The internal energy of an ideal gas is determined by its temperature and number of degrees of freedom. For a monatomic gas, $U = \frac{3}{2}nRT$, and for a diatomic gas $U = \frac{5}{2}nRT$.
		Formula/Rule:  $U_{mono} = \frac{3}{2}nRT$ and $U_{diatomic} = \frac{5}{2}nRT$
	2.	Calculate the total initial internal energy.
		Working: $U_{initial} = U_{1} + U_{2} = \frac{3}{2}n_1RT_1 + \frac{5}{2}n_2RT_2 = \frac{3}{2}(2)RT_1 + \frac{5}{2}(1)RT_2 = 3RT_1 + \frac{5}{2}RT_2$
		Key Point:  Since there is no heat exchange with the surroundings, the total internal energy remains constant.
	3.	Calculate the total final internal energy.
		Working:  Let the final temperature be $T_f$. The total number of moles is $n = n_1 + n_2 = 2+1 = 3$. The final internal energy is $U_{final} = \frac{3}{2}n_1RT_f + \frac{5}{2}n_2RT_f$. Substituting $n_1=2$ and $n_2=1$, we have a mixture, and we can't use the same formula. We need to take the average specific heat capacity. We use $U_{final} = n_{total}C_vT_f$ where $C_v = \frac{n_1C_{v1} + n_2C_{v2}}{n_1+n_2}$ and for a monoatomic $C_{v1} = \frac{3}{2}R$ and for a diatomic $C_{v2} = \frac{5}{2}R$. So, $C_v = \frac{2(\frac{3}{2}R) + 1(\frac{5}{2}R)}{3} = \frac{3R+\frac{5}{2}R}{3} = \frac{11R}{6}$. Hence, $U_{final} = 3(\frac{11}{6}R)T_f = \frac{11}{2}RT_f$
	4.	Equate the initial and final internal energies and solve for $T_f$.
		Working:  $U_{initial} = U_{final}$, so $3RT_1 + \frac{5}{2}RT_2 = \frac{11}{2}RT_f $. Dividing by $R$ and multiplying by 2, we have $6T_1 + 5T_2 = 11T_f$. Therefore, $T_f = \frac{6T_1 + 5T_2}{11}$
		Result: $T_f = \frac{6T_1 + 5T_2}{11}$
		Verification: The final temperature is a weighted average of the initial temperatures, which makes sense.
 ## OPTION ANALYSIS
[A] :x: This option is incorrect. It would be correct if both gases were monatomic, which is not the case here.
[B] ✓: This option is correct. It matches the final temperature derived in the solution.
[C] :x: This option is incorrect. The coefficients are wrong and the weighted average is not correct.
[D] :x: This option is incorrect. It is a simple average of the temperatures and does not take into account the number of moles and degrees of freedom of each gas.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Assuming the final temperature is a simple average of the initial temperatures. This is incorrect because the gases have different heat capacities and amounts.
		2.[Error 2]: Incorrectly calculating the internal energy of diatomic or monatomic gas. Use the correct formulas for $U$ for each type of gas.
	**Quick Tips**:
		1.[Helpful shortcut]: Remember the internal energy change is zero in an insulated container. Equate initial and final internal energies.
		2.[Memory aid]: Monatomic gas has 3 degrees of freedom, diatomic gas has 5 degrees of freedom.

VENKATESWARARAO · Answer

$\frac{2T_1 + T_2}{3}$

VENKATESWARARAO · Answer

The correct Option is "$\frac{6T_1 + 5T_2}{11}$"

VENKATESWARARAO · Answer

$\frac{3T_1 + 5T_2}{8}$

VENKATESWARARAO · Answer

$\frac{T_1 + T_2}{2}$

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