Calculate difference in specific heats for `1` gram of air at `N.T.P`. Given density of air at N.T.P. is `1.293 g litre^(-1), j=4.2xx10^(7)" erg "cal^(-1)`.

Explanation

To calculate the difference in specific heats (Cp - Cv) for 1 gram of air at Normal Temperature and Pressure (NTP), we will follow these steps:

### Step 1: Understand the relationship between Cp, Cv, and R
The difference between the specific heats of an ideal gas can be expressed as:
\[ Cp - Cv = R \]
Where R is the gas constant.

### Step 2: Convert the given density of air to appropriate units
The density of air at NTP is given as:
\[ \text{Density} = 1.293 \, \text{g/L} \]
We need to convert this to grams per cubic centimeter (g/cm³):
\[ 1 \, \text{L} = 1000 \, \text{cm}^3 \]
Thus,
\[ \text{Density} = \frac{1.293 \, \text{g}}{1000 \, \text{cm}^3} = 1.293 \times 10^{-3} \, \text{g/cm}^3 \]

### Step 3: Use the ideal gas equation to find R
Using the ideal gas equation:
\[ PV = nRT \]
For 1 gram of air (n = 1 g), we can express R as:
\[ R = \frac{PV}{T} \]
Where:
- P (pressure at NTP) = \( 1013 \times 10^6 \, \text{dyn/cm}^2 \) (since 1 atm = \( 1.013 \times 10^6 \, \text{dyn/cm}^2 \))
- V (volume) can be calculated from density:
\[ V = \frac{m}{\text{Density}} = \frac{1 \, \text{g}}{1.293 \times 10^{-3} \, \text{g/cm}^3} \approx 773.5 \, \text{cm}^3 \]
- T (temperature at NTP) = 273 K

### Step 4: Calculate R
Now substituting the values into the equation for R:
\[ R = \frac{P \cdot V}{T} = \frac{(1013 \times 10^6) \cdot (773.5)}{273} \]

### Step 5: Convert R to calories
To convert R from ergs to calories, we use the conversion:
\[ 1 \, \text{cal} = 4.2 \times 10^7 \, \text{ergs} \]
Thus,
\[ R \text{ (in calories)} = \frac{R \text{ (in ergs)}}{4.2 \times 10^7} \]

### Step 6: Calculate Cp - Cv
Finally, since \( Cp - Cv = R \), we can find the difference in specific heats:
\[ Cp - Cv = R \text{ (in calories)} \]

### Final Calculation
Perform the calculations to find the numerical value of \( Cp - Cv \).

### Answer
After performing the calculations, we find:
\[ Cp - Cv \approx 0.068 \, \text{cal/g/K} \]