Question

Two ideal gases at absolute temperature \(T_1\) and \(T_2\) are mixed. In this process there is no loss of energy. If the molecular masses are \(m_1\) and \(m_2\), and mole number of the gases are \(n_1\) and \(n_2\), then the final temperature of the mixture is:

• A. \( \frac{n_1T_1 + n_2T_2}{n_1 + n_2} \)
• B. \( \frac{m_1T_1 + m_2T_2}{n_1 + n_2} \)
• C. \( \frac{n_1T_1 + n_2T_2}{m_1 + m_2} \)
• D. \( T = \sqrt{T_1 + T_2} \)

Hint:

When mixing gases, the final temperature can be calculated by the weighted average of the initial temperatures, weighted by the number of moles of each gas.

Answer 1

The Correct Option is A

When two ideal gases at absolute temperatures \( T_1 \) and \( T_2 \) are mixed, the total energy is conserved, and there is no loss of energy in the process. The total internal energy of the system is the sum of the internal energies of the individual gases.

The internal energy of an ideal gas is proportional to its temperature, and for each gas, we have:

\[ U_1 = n_1 C_V T_1 \] \[ U_2 = n_2 C_V T_2 \]

where:

• \( n_1 \) and \( n_2 \) are the number of moles of gases 1 and 2,
• \( C_V \) is the molar specific heat at constant volume (which is constant for both gases),
• \( T_1 \) and \( T_2 \) are the absolute temperatures of the two gases.

When the two gases are mixed, the total internal energy becomes:

\[ U_{\text{total}} = U_1 + U_2 = n_1 C_V T_1 + n_2 C_V T_2 \]

The final temperature \( T \) is such that the total energy is divided between the two gases. So, we have:

\[ U_{\text{total}} = (n_1 + n_2) C_V T \]

Equating the total energy before and after mixing:

\[ n_1 C_V T_1 + n_2 C_V T_2 = (n_1 + n_2) C_V T \]

Simplifying:

\[ n_1 T_1 + n_2 T_2 = (n_1 + n_2) T \]

Thus, the final temperature \( T \) is:

\[ T = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2} \]

Therefore, the correct answer is Option (1), \( T = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2} \).

Answer 2

To solve the problem, we need to determine the final temperature of the mixture of two ideal gases that are mixed at absolute temperatures \( T_1 \) and \( T_2 \) without any loss of energy.

1. Understanding the Problem:
We are given:

• Two ideal gases with absolute temperatures \( T_1 \) and \( T_2 \).
• The molecular masses of the gases are \( m_1 \) and \( m_2 \), and the number of moles of the gases are \( n_1 \) and \( n_2 \).
• We need to find the final temperature \( T \) of the mixture when the gases are mixed and no energy is lost.

2. Applying the Principle of Conservation of Energy:
Since no energy is lost in the process, the total energy of the system remains constant. The internal energy of each gas is proportional to its temperature and the number of moles. The final temperature of the mixture can be found by taking the weighted average of the temperatures of the individual gases based on their mole numbers: \[ T = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2} \] This equation gives the final temperature as a function of the initial temperatures and the number of moles of the two gases.

Final Answer:
The correct option is (A) \( T = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2} \).