Calculate the work done by the system in an irreversible (single step) adiabatic expansion of 2 mole of a polyatomic gas `(gamma=4//3)` from 300 K and pressure 10 atm to 1 atm :

Options

A. `-227`
B. `-205 R`
C. `-405 R`
D. None of these

Explanation

To calculate the work done by the system in an irreversible adiabatic expansion of 2 moles of a polyatomic gas (with \(\gamma = \frac{4}{3}\)) from an initial state of 300 K and 10 atm to a final pressure of 1 atm, we can follow these steps:

### Step 1: Understand the Process
In an irreversible adiabatic expansion, the heat exchange (\(q\)) is zero. Therefore, the work done (\(W\)) by the system can be expressed as:
\[ W = \Delta U \]
where \(\Delta U\) is the change in internal energy.

### Step 2: Calculate Change in Internal Energy
The change in internal energy for an ideal gas can be calculated using the formula:
\[ \Delta U = n C_v \Delta T \]
where:
- \(n\) = number of moles
- \(C_v\) = molar heat capacity at constant volume
- \(\Delta T = T_2 - T_1\)

### Step 3: Find \(C_v\) for Polyatomic Gas
For a polyatomic gas, the molar heat capacity at constant volume is given by:
\[ C_v = \frac{3R}{2} \text{ (for diatomic gas, but for polyatomic gas, it is } 3R) \]

### Step 4: Find Final Temperature \(T_2\)
To find \(T_2\), we can use the ideal gas law and the fact that the process is adiabatic. We know:
- Initial pressure \(P_1 = 10 \, \text{atm}\)
- Final pressure \(P_2 = 1 \, \text{atm}\)
- Initial temperature \(T_1 = 300 \, \text{K}\)

Using the relation for an adiabatic process:
\[ \frac{P_1 V_1^\gamma}{T_1} = \frac{P_2 V_2^\gamma}{T_2} \]
However, we can also use the relation:
\[ \frac{T_2}{T_1} = \left(\frac{P_1}{P_2}\right)^{\frac{\gamma - 1}{\gamma}} \]
Substituting the values:
\[ \frac{T_2}{300} = \left(\frac{10}{1}\right)^{\frac{\frac{4}{3} - 1}{\frac{4}{3}}} = 10^{\frac{1}{4}} \approx 1.778 \]
Thus,
\[ T_2 \approx 300 \times 1.778 \approx 533.4 \, \text{K} \]

### Step 5: Calculate \(\Delta T\)
Now we can find \(\Delta T\):
\[ \Delta T = T_2 - T_1 = 533.4 - 300 = 233.4 \, \text{K} \]

### Step 6: Calculate Work Done
Now substituting into the equation for \(\Delta U\):
\[ \Delta U = n C_v \Delta T = 2 \times 3R \times 233.4 \]
\[ \Delta U = 6R \times 233.4 = 1400.4R \]
Thus, the work done \(W\) is:
\[ W = -\Delta U = -1400.4R \]

### Final Answer
The work done by the system in the irreversible adiabatic expansion is:
\[ W = -1400.4R \, \text{(in terms of R)} \]