The relation between U, P and V for an iodeal gas is U=2+3PV. What is the atomicity of the gas.

Options

A. monoatomic
B. diatomic
C. polyatomic
D. either a monoatomic or diatomic

Answer

To find the atomicity of the gas given the relation \( U = 2 + 3PV \), we can follow these steps:

### Step 1: Understand the given relation
The equation \( U = 2 + 3PV \) relates the internal energy \( U \) of the gas to its pressure \( P \) and volume \( V \). Here, \( U \) is the internal energy, and \( PV \) is the product of pressure and volume.

### Step 2: Apply the First Law of Thermodynamics
According to the First Law of Thermodynamics, we have:
\[ dQ = dU + dW \]
For an adiabatic process, the heat exchange \( dQ = 0 \), which implies:
\[ dU + dW = 0 \quad \Rightarrow \quad dU = -dW \]

### Step 3: Differentiate the internal energy equation
We differentiate \( U \) with respect to \( P \) and \( V \):
\[ dU = \frac{\partial U}{\partial P} dP + \frac{\partial U}{\partial V} dV \]
Calculating the partial derivatives from \( U = 2 + 3PV \):
- The derivative with respect to \( V \) is:
\[ \frac{\partial U}{\partial V} = 3P \]
- The derivative with respect to \( P \) is:
\[ \frac{\partial U}{\partial P} = 3V \]
Thus, we have:
\[ dU = 3P dV + 3V dP \]

### Step 4: Express the work done \( dW \)
The work done \( dW \) in a thermodynamic process is given by:
\[ dW = P dV \]
Substituting this into the equation \( dU = -dW \):
\[ 3P dV + 3V dP = -P dV \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ 3P dV + P dV + 3V dP = 0 \quad \Rightarrow \quad (4P dV + 3V dP) = 0 \]

### Step 6: Separate the variables
We can separate the variables:
\[ \frac{4P}{3V} dV + dP = 0 \]
This can be rearranged to:
\[ \frac{dP}{P} = -\frac{4}{3} \frac{dV}{V} \]

### Step 7: Integrate both sides
Integrating both sides gives:
\[ \int \frac{dP}{P} = -\frac{4}{3} \int \frac{dV}{V} \]
This results in:
\[ \ln P = -\frac{4}{3} \ln V + C \]
Exponentiating both sides results in:
\[ P V^{\frac{4}{3}} = e^C \]
Let \( K = e^C \), then:
\[ PV^{\frac{4}{3}} = K \]

### Step 8: Compare with the general gas equation
The general form of the adiabatic process for an ideal gas is:
\[ PV^\gamma = \text{constant} \]
By comparing, we find:
\[ \gamma = \frac{4}{3} \]

### Step 9: Determine the atomicity
For ideal gases, the value of \( \gamma \) helps us determine the atomicity:
- For monatomic gases, \( \gamma = \frac{5}{3} \)
- For diatomic gases, \( \gamma = \frac{7}{5} \)
- For polyatomic gases, \( \gamma < \frac{5}{3} \)

Since \( \gamma = \frac{4}{3} \), this indicates that the gas is polyatomic.

### Conclusion
The atomicity of the gas is 4, as it is a polyatomic gas.

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