Question

0.5 mole of \( {N}_2 \) at 27°C is mixed with 0.5 mole of \( {O}_2 \) at 42°C. The temperature of the mixture is:

• A. 42°C
• B. 34.5°C
• C. 32.5°C
• D. 37.5°C
• E. 27°C

Hint:

To find the final temperature when mixing substances, use the conservation of energy principle, assuming no heat loss to the surroundings.

Answer 1

The Correct Option is B

When two bodies at different temperatures are mixed, the final temperature \( T_f \) of the mixture can be calculated using the principle of conservation of energy. The heat gained by the colder body is equal to the heat lost by the hotter body. 
The equation for this can be written as: \[ m_1 C_1 (T_f - T_1) = m_2 C_2 (T_2 - T_f) \] Where: - \( m_1 \) and \( m_2 \) are the masses of the two substances,
- \( C_1 \) and \( C_2 \) are their specific heats (for simplicity, assume both gases have similar specific heat values),
- \( T_1 \) and \( T_2 \) are their initial temperatures,
- \( T_f \) is the final temperature.
Since we are given that both substances are in equal moles (0.5 mole each), and assuming similar specific heat values for \( {N}_2 \) and \( {O}_2 \), we can simplify the equation by setting the specific heats and masses to equal values. 
Thus, the equation simplifies to: \[ (T_f - 27) = (42 - T_f) \] Solving for \( T_f \): \[ T_f - 27 = 42 - T_f 
2T_f = 69 
T_f = 34.5°C \] Thus, the final temperature of the mixture is \( 34.5°C \).
Therefore, the correct answer is option (B), 34.5°C.