Given, for air `c_(upsilon) =0.162 cal g^(-1) K^(-1)` and density at NTP is `0.001293 g cm^(-3)` What is the value of `c_p ?`

Options

A. `0.123 cal g^(-1) K^(-1)`
B. `0.23 cal g^(-1) K^(-1)`
C. `0.246 cal^(-1) g^(-1) K^(-1)`
D. `0.46 cal^(-1) g^(-1)`

Answer

To find the value of \( c_p \) for air given \( c_v \) and the density at NTP, we can follow these steps:

### Step 1: Understand the relationship between \( c_p \) and \( c_v \)
The relationship between the specific heats at constant pressure (\( c_p \)) and constant volume (\( c_v \)) is given by:
\[ c_p - c_v = R \]
where \( R \) is the specific gas constant.

### Step 2: Calculate the specific gas constant \( R \)
To find \( R \), we can use the equation:
\[ R = \frac{P V}{n T} \]
For one mole of gas, this can be simplified to:
\[ R = \frac{P}{\rho J T} \]
where:
- \( P \) is the pressure,
- \( \rho \) is the density,
- \( J \) is the mechanical equivalent of heat (approximately \( 4.2 \, \text{J/g°C} \)),
- \( T \) is the absolute temperature in Kelvin.

### Step 3: Substitute the known values
Given:
- \( c_v = 0.162 \, \text{cal/g K} \)
- Density \( \rho = 0.001293 \, \text{g/cm}^3 = 1.293 \, \text{kg/m}^3 \)
- Pressure at NTP \( P = 1.01 \times 10^5 \, \text{Pa} \)
- Temperature at NTP \( T = 273 \, \text{K} \)

Now, we can calculate \( R \):
\[ R = \frac{P}{\rho J T} \]
Substituting the values:
\[ R = \frac{1.01 \times 10^5 \, \text{Pa}}{(1.293 \, \text{kg/m}^3)(4.2 \, \text{J/g°C})(273 \, \text{K})} \]

### Step 4: Calculate \( R \)
First, convert \( J \) to \( \text{cal} \):
\[ 1 \, \text{cal} = 4.184 \, \text{J} \]
Thus, \( J \) in terms of calories is:
\[ J = \frac{4.2}{4.184} \, \text{cal/g°C} \approx 1 \, \text{cal/g°C} \]
Now substituting:
\[ R = \frac{1.01 \times 10^5}{(1.293)(1)(273)} \]
Calculating the denominator:
\[ R = \frac{1.01 \times 10^5}{352.659} \approx 286.5 \, \text{J/kg K} \]

### Step 5: Convert \( R \) to cal
To convert \( R \) from J to cal:
\[ R \approx \frac{286.5}{4.184} \approx 68.5 \, \text{cal/kg K} \]

### Step 6: Calculate \( c_p \)
Now substituting back into the equation \( c_p = c_v + R \):
\[ c_p = 0.162 + 0.0685 \approx 0.2305 \, \text{cal/g K} \]

### Step 7: Final answer
Thus, the value of \( c_p \) is approximately:
\[ c_p \approx 0.23 \, \text{cal/g K} \]