14 g of CO at `27^@ C` is mixed with 16'g of `O_2` at `47^@ C`. The temperature of mixture is (vibartion mode neglected)

Options

A. `-5^@ C`
B. `32^@ C`
C. `37^@ C`
D. `27^@ C`

Answer

To solve the problem of finding the temperature of the mixture of 14 g of CO at 27°C and 16 g of O2 at 47°C, we will use the concept of internal energy and the conservation of energy principle. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the number of moles of CO and O2
- **For CO:**
- Molar mass of CO = 28 g/mol
- Mass of CO = 14 g
- Number of moles of CO (n1) = mass / molar mass = 14 g / 28 g/mol = 0.5 moles

- **For O2:**
- Molar mass of O2 = 32 g/mol
- Mass of O2 = 16 g
- Number of moles of O2 (n2) = mass / molar mass = 16 g / 32 g/mol = 0.5 moles

### Step 2: Determine the degrees of freedom (F) for both gases
- Both CO and O2 are diatomic gases, which means they have 5 degrees of freedom (3 translational + 2 rotational).
- Therefore, F = 5 for both CO and O2.

### Step 3: Write the expression for internal energy (U)
- The internal energy (U) for an ideal gas is given by the formula:
\[ U = \frac{F}{2} nRT \]
- For CO:
\[ U_{CO} = \frac{5}{2} n_1 R T_1 = \frac{5}{2} \times 0.5 \times R \times 300 \text{ K} \]
- For O2:
\[ U_{O2} = \frac{5}{2} n_2 R T_2 = \frac{5}{2} \times 0.5 \times R \times 320 \text{ K} \]

### Step 4: Set up the equation for conservation of internal energy
- The total internal energy of the mixture after mixing will be equal to the sum of the internal energies of CO and O2:
\[ U_{total} = U_{CO} + U_{O2} \]
- Let T be the final temperature of the mixture. Then:
\[ U_{total} = \frac{5}{2} (n_1 + n_2) R T \]
- Substituting the values:
\[ \frac{5}{2} \times (0.5 + 0.5) R T = \frac{5}{2} \times 0.5 R \times 300 + \frac{5}{2} \times 0.5 R \times 320 \]

### Step 5: Simplify and solve for T
- The left side simplifies to:
\[ 5 R T \]
- The right side simplifies to:
\[ \frac{5}{2} R (300 + 320) = \frac{5}{2} R \times 620 \]
- Setting the two sides equal:
\[ 5 R T = \frac{5}{2} R \times 620 \]
- Dividing both sides by 5R:
\[ T = \frac{620}{2} = 310 \text{ K} \]

### Step 6: Convert the final temperature to Celsius
- To convert Kelvin to Celsius:
\[ T_{Celsius} = T_{Kelvin} - 273 = 310 - 273 = 37 \text{ °C} \]

### Final Answer:
The temperature of the mixture is **37°C**.