Question

If \(2\) mole of an ideal monoatomic gas at temperature \(T\) is mixed with \(6\) mole of another ideal monoatomic gas at temperature \(2T\) then the temperature of mixture is :

• A. \( \frac{5}{2}T \)
• B. \( \frac{5}{4}T \)
• C. \( \frac{7}{2}T \)
• D. \( \frac{7}{4}T \)

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem involves finding the final equilibrium temperature of a mixture of two monoatomic ideal gases.
Step 2: Key Formula or Approach:
By the principle of conservation of energy (in an insulated system), the total internal energy of the mixture is the sum of the initial internal energies of the individual gases.
For an ideal gas, \(U = n C_V T\). For monoatomic gases, \(C_V = \frac{3}{2} R\).
\[ (n_1 + n_2) C_V T_{mix} = n_1 C_V T_1 + n_2 C_V T_2 \]
Since \(C_V\) is the same for both: \(T_{mix} = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}\).
Step 3: Detailed Explanation:
Given:
\(n_1 = 2\), \(T_1 = T\)
\(n_2 = 6\), \(T_2 = 2T\)
Applying the formula:
\[ T_{mix} = \frac{(2 \times T) + (6 \times 2T)}{2 + 6} \]
\[ T_{mix} = \frac{2T + 12T}{8} = \frac{14T}{8} \]
\[ T_{mix} = \frac{7}{4} T \]
Step 4: Final Answer:
The resulting temperature of the mixture is \(\frac{7}{4} T\).