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 a mixture of gases, we will follow these steps:

### Step 1: Identify the Given Data
- Mass of CO = 14 g
- Temperature of CO (T1) = 27°C = 300 K (convert to Kelvin by adding 273)
- Mass of O2 = 16 g
- Temperature of O2 (T2) = 47°C = 320 K (convert to Kelvin by adding 273)

### Step 2: Calculate the Number of Moles
- Molecular mass of CO = 28 g/mol
- Number of moles of CO (n1) = Mass / Molar mass = 14 g / 28 g/mol = 0.5 mol

- Molecular mass of O2 = 32 g/mol
- Number of moles of O2 (n2) = Mass / Molar mass = 16 g / 32 g/mol = 0.5 mol

### Step 3: Total Number of Moles
- Total number of moles (N) = n1 + n2 = 0.5 mol + 0.5 mol = 1 mol

### Step 4: Use the Formula for Internal Energy
The internal energy (U) of an ideal gas can be expressed as:
\[ U = \frac{F}{2} nRT \]
Where:
- F = degrees of freedom (for diatomic gases, F = 5)
- n = number of moles
- R = universal gas constant (which will cancel out later)
- T = absolute temperature in Kelvin

### Step 5: Set Up the Equation for Internal Energy Conservation
The total internal energy of the mixture (Ut) is equal to the sum of the internal energies of CO and O2:
\[ U_t = U_{CO} + U_{O_2} \]

Substituting the formula for internal energy:
\[ U_t = \frac{F}{2} n_1 R T_1 + \frac{F}{2} n_2 R T_2 \]

### Step 6: Substitute Values
Since both gases have the same degrees of freedom (F = 5) and R will cancel out, we can simplify:
\[ U_t = \frac{5}{2} (0.5) R (300) + \frac{5}{2} (0.5) R (320) \]

### Step 7: Simplify the Equation
\[ U_t = \frac{5R}{4} (300 + 320) \]
\[ U_t = \frac{5R}{4} (620) \]

### Step 8: Find the Final Temperature
Using the total number of moles (N = 1):
\[ U_t = \frac{5}{2} (1) R T \]
Setting the two expressions for Ut equal to each other:
\[ \frac{5R}{4} (620) = \frac{5}{2} (1) R T \]

### Step 9: Cancel R and Solve for T
\[ \frac{5}{4} (620) = \frac{5}{2} T \]
\[ 620 = 2T \]
\[ T = 310 \, K \]

### Step 10: Convert to Celsius
To convert Kelvin to Celsius:
\[ T_{Celsius} = T_{Kelvin} - 273 \]
\[ T_{Celsius} = 310 - 273 = 37 \, °C \]

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

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