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), \( T = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2} \).