The density of a polyatomic gas in stantard conditions is `0.795 kg m^(-3)`. The specific heat of the gas at constant

Options

A. `930 J-kg^(-1) K^(-1)`
B. `1400 J-kg^(-1) K^(-1)`
C. `1120 J-kg^(-1) K^(-1)`
D. `1600 J-kg^(-1) K^(-1)`

Explanation

To solve the problem step by step, we will follow the approach outlined in the video transcript.

### Step 1: Understand the Given Data
We are given the density of a polyatomic gas at standard conditions:
- Density (ρ) = 0.795 kg/m³
- Standard pressure (P) = 1 atm = 10^5 Pa
- Standard temperature (T) = 273 K

### Step 2: Use the Ideal Gas Equation
The ideal gas equation can be expressed in terms of mass:
\[ PV = mRT \]
Where:
- \( P \) = Pressure
- \( V \) = Volume
- \( m \) = Mass
- \( R \) = Specific gas constant
- \( T \) = Temperature

### Step 3: Rearrange the Ideal Gas Equation
We can rearrange the equation to find the specific gas constant \( R \):
\[ R = \frac{PV}{mT} \]

### Step 4: Relate Mass and Density
We know that density \( \rho \) is defined as:
\[ \rho = \frac{m}{V} \]
From this, we can express volume \( V \) in terms of mass \( m \) and density \( \rho \):
\[ V = \frac{m}{\rho} \]

### Step 5: Substitute Volume in the Ideal Gas Equation
Substituting \( V \) in the equation for \( R \):
\[ R = \frac{P \cdot \frac{m}{\rho}}{mT} \]
This simplifies to:
\[ R = \frac{P}{\rho T} \]

### Step 6: Substitute Known Values
Now we can substitute the known values into the equation:
\[ R = \frac{10^5 \, \text{Pa}}{0.795 \, \text{kg/m}^3 \cdot 273 \, \text{K}} \]

### Step 7: Calculate \( R \)
Calculating the value:
\[ R = \frac{10^5}{0.795 \cdot 273} \]
\[ R = \frac{10^5}{216.795} \]
\[ R \approx 460.7 \, \text{J/(kg K)} \]

### Step 8: Determine Specific Heat at Constant Volume
For a polyatomic gas, the specific heat at constant volume \( C_v \) can be calculated using the formula:
\[ C_v = \frac{R}{\gamma - 1} \]
Where \( \gamma \) (gamma) for a polyatomic gas is approximately \( \frac{5}{3} \).

### Step 9: Calculate \( C_v \)
Substituting \( R \) and \( \gamma \):
\[ C_v = \frac{460.7}{\frac{5}{3} - 1} \]
\[ C_v = \frac{460.7}{\frac{2}{3}} \]
\[ C_v = 460.7 \cdot \frac{3}{2} \]
\[ C_v \approx 691.05 \, \text{J/(kg K)} \]

### Step 10: Check the Options
The calculated value of \( C_v \) does not match the options provided. However, if we consider the specific heat at constant pressure \( C_p \) instead, we can use:
\[ C_p = C_v + R \]
\[ C_p = 691.05 + 460.7 \]
\[ C_p \approx 1151.75 \, \text{J/(kg K)} \]

This value still does not match any of the options.

### Conclusion
The specific heat at constant volume \( C_v \) is approximately 691.05 J/(kg K) and the specific heat at constant pressure \( C_p \) is approximately 1151.75 J/(kg K). The closest option might be considered based on the context of the question.