Two moles of gas A at `27^(@)C` mixed with a 3 moles of gas at `37^(@)C.` If both are monatomic ideal gases, what will be the temperature of the mixture ?

Options

A. `66^(@)C`
B. `11^(@)C`
C. `22^(@)C`
D. `33^(@)C`

Explanation

To find the temperature of the mixture of two gases, we can use the formula for the weighted average temperature based on the number of moles of each gas. The formula is:

\[ T = \frac{\mu_1 \cdot T_1 + \mu_2 \cdot T_2}{\mu_1 + \mu_2} \]

Where:
- \( T \) is the temperature of the mixture.
- \( \mu_1 \) and \( \mu_2 \) are the number of moles of gas A and gas B, respectively.
- \( T_1 \) and \( T_2 \) are the temperatures of gas A and gas B, respectively.

### Step 1: Convert temperatures to Kelvin
First, we need to convert the given temperatures from Celsius to Kelvin:

- For gas A at \( 27^\circ C \):
\[ T_1 = 27 + 273 = 300 \, K \]

- For gas B at \( 37^\circ C \):
\[ T_2 = 37 + 273 = 310 \, K \]

### Step 2: Identify the number of moles
From the problem, we know:
- \( \mu_1 = 2 \) moles (for gas A)
- \( \mu_2 = 3 \) moles (for gas B)

### Step 3: Substitute values into the formula
Now we can substitute the values into the formula:

\[ T = \frac{(2 \cdot 300) + (3 \cdot 310)}{2 + 3} \]

### Step 4: Calculate the numerator
Calculate the numerator:
\[ 2 \cdot 300 = 600 \]
\[ 3 \cdot 310 = 930 \]
\[ \text{Numerator} = 600 + 930 = 1530 \]

### Step 5: Calculate the denominator
Calculate the denominator:
\[ \mu_1 + \mu_2 = 2 + 3 = 5 \]

### Step 6: Calculate the temperature of the mixture
Now, substitute the numerator and denominator back into the equation:
\[ T = \frac{1530}{5} = 306 \, K \]

### Step 7: Convert back to Celsius
Finally, convert the temperature back to Celsius:
\[ T_{Celsius} = 306 - 273 = 33^\circ C \]

### Final Answer
The temperature of the mixture is \( 33^\circ C \).

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