2 moles of an ideal monoatomic gas undergo a reversible process for which `PV^(2)=` constant. The gas sample is made to expand from initial volume of 1 litre of final volume of 3 litre starting from initial temperature of 300 K. find the value of `DeltaS_("sys")` for the above process. Report your answer as `Y` where `DeltaS_("sys")=-YRIn3`

Answer

To find the change in entropy (ΔS_sys) for the given process of an ideal monoatomic gas, we will follow these steps:

### Step 1: Understand the Process
We are given that the process follows the equation \( PV^2 = \text{constant} \). For an ideal gas, we also know that \( PV = nRT \).

### Step 2: Relate Temperature and Volume
From the equation \( PV^2 = \text{constant} \), we can express pressure \( P \) in terms of volume \( V \):
\[ P = \frac{\text{constant}}{V^2} \]
Substituting \( P \) from the ideal gas equation:
\[ \frac{nRT}{V} \cdot V^2 = \text{constant} \]
This simplifies to:
\[ TV = \text{constant} \]
Thus, we can say that:
\[ T_1 V_1 = T_2 V_2 \]

### Step 3: Calculate the Temperature Ratio
Given:
- Initial volume \( V_1 = 1 \, \text{L} \)
- Final volume \( V_2 = 3 \, \text{L} \)
- Initial temperature \( T_1 = 300 \, \text{K} \)

Using the relation \( T_1 V_1 = T_2 V_2 \):
\[ T_2 = \frac{T_1 V_1}{V_2} = \frac{300 \, \text{K} \cdot 1 \, \text{L}}{3 \, \text{L}} = 100 \, \text{K} \]

### Step 4: Calculate Change in Entropy (ΔS_sys)
The change in entropy for a reversible process can be calculated using:
\[ \Delta S = n C_m \ln \frac{T_2}{T_1} \]
For a monoatomic ideal gas, the molar specific heat \( C_m \) for a polytropic process can be calculated as:
\[ C_m = \frac{R}{2} \]
Thus, substituting in the values:
\[ \Delta S = n \cdot \frac{R}{2} \ln \frac{T_2}{T_1} \]
Substituting \( n = 2 \) moles, \( T_1 = 300 \, \text{K} \), and \( T_2 = 100 \, \text{K} \):
\[ \Delta S = 2 \cdot \frac{R}{2} \ln \frac{100}{300} = R \ln \frac{1}{3} = -R \ln 3 \]

### Step 5: Report the Value of Y
According to the problem statement, we need to express the result as:
\[ \Delta S_{\text{sys}} = -Y R \ln 3 \]
From our calculation, we have:
\[ \Delta S_{\text{sys}} = -R \ln 3 \]
Thus, comparing, we find:
\[ Y = 1 \]

### Final Answer
The value of \( Y \) is \( 1 \).

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