Find the specific heat of a polyatomic gas at constant volume if the density of this gas in standard conditions is `7.95 xx 10^(-4) gm//cm^3`. Express your result in cal/gm degree.

Answer

To find the specific heat of a polyatomic gas at constant volume, we can use the formula:

\[ C_v = \frac{R}{\gamma - 1} \]

where:
- \( C_v \) is the specific heat at constant volume,
- \( R \) is the universal gas constant,
- \( \gamma \) is the ratio of specific heats \( C_p/C_v \).

For polyatomic gases, \( \gamma \) typically ranges from 1.3 to 1.4. However, we will use a common value of \( \gamma = \frac{7}{5} = 1.4 \) for our calculations.

### Step 1: Determine the density of the gas
The density of the gas is given as \( 7.95 \times 10^{-4} \, \text{gm/cm}^3 \).

### Step 2: Convert density to kg/m³
To convert the density from gm/cm³ to kg/m³, we use the conversion factor \( 1 \, \text{gm/cm}^3 = 1000 \, \text{kg/m}^3 \).

\[ \text{Density} = 7.95 \times 10^{-4} \, \text{gm/cm}^3 \times 1000 \, \text{kg/m}^3 = 0.795 \, \text{kg/m}^3 \]

### Step 3: Use the ideal gas law to find the molar mass
The ideal gas law is given by:

\[ PV = nRT \]

Where:
- \( P \) is the pressure (standard conditions: \( P = 1 \, \text{atm} = 101325 \, \text{Pa} \)),
- \( V \) is the volume,
- \( n \) is the number of moles,
- \( R \) is the universal gas constant (\( R = 8.314 \, \text{J/(mol K)} \)),
- \( T \) is the temperature (standard conditions: \( T = 273.15 \, \text{K} \)).

To find the molar mass \( M \), we can rearrange the ideal gas law:

\[ M = \frac{dRT}{P} \]

Substituting the values:

\[ M = \frac{(0.795 \, \text{kg/m}^3)(8.314 \, \text{J/(mol K)})(273.15 \, \text{K})}{101325 \, \text{Pa}} \]

Calculating this gives:

\[ M \approx \frac{(0.795)(8.314)(273.15)}{101325} \approx 0.018 \, \text{kg/mol} \approx 18 \, \text{g/mol} \]

### Step 4: Calculate the specific heat at constant volume
Now we can substitute the values into the specific heat formula:

\[ C_v = \frac{R}{\gamma - 1} \]

Using \( R = 1.987 \, \text{cal/(mol K)} \) (since we need the answer in cal/gm degree) and \( \gamma = 1.4 \):

\[ C_v = \frac{1.987}{1.4 - 1} = \frac{1.987}{0.4} \approx 4.9675 \, \text{cal/(mol K)} \]

### Step 5: Convert to cal/gm degree
To convert \( C_v \) from cal/(mol K) to cal/gm degree, we divide by the molar mass in grams:

\[ C_v = \frac{4.9675 \, \text{cal/(mol K)}}{18 \, \text{g/mol}} \approx 0.276 \, \text{cal/g K} \]

### Final Answer
The specific heat of the polyatomic gas at constant volume is approximately \( 0.276 \, \text{cal/g K} \).

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