If the ratio of the specific heats of steam is 1.33 and R=8312 J/k mole k find the molar heat capacities of steam at constant pressure and constant volume.

Options

A. 33.5 kJ/k mole.
B. 25.19 kJ/kg K
C. 25.19 kJ/K mole.
D. 24.12 kj/k mole 16.12kj/k mole

Explanation

To find the molar heat capacities of steam at constant pressure (C_p) and constant volume (C_v), we can use the relationship between the specific heats and the gas constant (R). Given that the ratio of the specific heats (γ) is 1.33, we can use the following equations:

1. **Relationship between C_p, C_v, and R:**
\[ C_p - C_v = R \]

2. **Relationship between C_p, C_v, and the ratio of specific heats (γ):**
\[ \gamma = \frac{C_p}{C_v} \]

### Step 1: Express C_p in terms of C_v
From the second equation, we can express C_p as:
\[ C_p = \gamma C_v \]

### Step 2: Substitute C_p in the first equation
Now, substitute C_p in the first equation:
\[ \gamma C_v - C_v = R \]

### Step 3: Factor out C_v
This simplifies to:
\[ C_v(\gamma - 1) = R \]

### Step 4: Solve for C_v
Now, we can solve for C_v:
\[ C_v = \frac{R}{\gamma - 1} \]

### Step 5: Plug in the values
Given R = 8312 J/(kmole·K) and γ = 1.33, we can substitute these values into the equation:
\[ C_v = \frac{8312}{1.33 - 1} \]
\[ C_v = \frac{8312}{0.33} \]
\[ C_v \approx 25100 \, \text{J/(kmole·K)} \]

### Step 6: Calculate C_p
Now that we have C_v, we can find C_p using the relationship from Step 1:
\[ C_p = \gamma C_v \]
\[ C_p = 1.33 \times 25100 \]
\[ C_p \approx 33353 \, \text{J/(kmole·K)} \]

### Final Results
- Molar heat capacity at constant volume (C_v) ≈ 25100 J/(kmole·K)
- Molar heat capacity at constant pressure (C_p) ≈ 33353 J/(kmole·K)