A sample of certain mas s of an ideal polyatomic gas is expanded against constant pressure of 1 atm adiabatically from volume 2L , pressure 6 atm and temperature 300K to state where its final volume is 8L . Then calculate entropy change (in J/k) in the process.(Neglect vibrational degress of freedom)[1 L atm = 100 J ,log 2 =0.3 , log 3 = 0.48 , log e = 2.3] (approximate integer)

Explanation

To calculate the entropy change (ΔS) for the given adiabatic expansion of an ideal polyatomic gas, we can follow these steps:

### Step 1: Identify the Initial and Final States
- Initial Volume (V1) = 2 L
- Final Volume (V2) = 8 L
- Initial Pressure (P1) = 6 atm
- External Pressure (P_ext) = 1 atm
- Initial Temperature (T1) = 300 K

### Step 2: Calculate the Work Done
The work done (W) during an irreversible process against constant external pressure can be calculated using the formula:
\[ W = -P_{\text{ext}} \Delta V \]
Where:
- \(\Delta V = V_2 - V_1 = 8 L - 2 L = 6 L\)

Substituting the values:
\[ W = -1 \, \text{atm} \times 6 \, \text{L} = -6 \, \text{L atm} \]

### Step 3: Convert Work Done to Joules
Using the conversion \(1 \, \text{L atm} = 100 \, \text{J}\):
\[ W = -6 \, \text{L atm} \times 100 \, \text{J/L atm} = -600 \, \text{J} \]

### Step 4: Calculate the Change in Internal Energy (ΔU)
For an ideal gas undergoing an adiabatic process:
\[ \Delta U = Q - W \]
Since the process is adiabatic, \(Q = 0\):
\[ \Delta U = -W = 600 \, \text{J} \]

### Step 5: Calculate the Final Pressure (P2)
Using the ideal gas law, we can find the final pressure (P2) after the expansion:
Using the relation:
\[ P_1 V_1 = nRT_1 \]
\[ P_2 V_2 = nRT_2 \]

Since the number of moles (n) and R are constant, we can write:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

We need to find \(T_2\) first. The ratio of temperatures can be derived from the pressures and volumes:
\[ \frac{T_2}{T_1} = \frac{P_1 V_1}{P_2 V_2} \]

### Step 6: Calculate the Temperature Ratio
Assuming \(P_2\) is unknown, we can express it in terms of known quantities:
Using the adiabatic relation for an ideal gas, we can also find:
\[ P_2 = P_1 \frac{V_1}{V_2} \]
Substituting the values:
\[ P_2 = 6 \, \text{atm} \times \frac{2 \, \text{L}}{8 \, \text{L}} = 1.5 \, \text{atm} \]

### Step 7: Calculate the Final Temperature (T2)
Using the ideal gas law:
\[ T_2 = \frac{P_2 V_2}{nR} \]
Since \(nR\) is constant, we can find the ratio:
\[ \frac{T_2}{T_1} = \frac{P_2 V_2}{P_1 V_1} \]
Substituting the known values:
\[ \frac{T_2}{300} = \frac{1.5 \times 8}{6 \times 2} = 2 \]
Thus, \(T_2 = 600 \, \text{K}\).

### Step 8: Calculate the Entropy Change (ΔS)
The entropy change for the process can be calculated using the formula:
\[ \Delta S = nC_V \ln \frac{T_2}{T_1} + nR \ln \frac{V_2}{V_1} \]

For a polyatomic gas, \(C_V = \frac{3}{2}R\) (considering 3 translational and 2 rotational degrees of freedom):
\[ \Delta S = n \left( \frac{3}{2}R \ln \frac{600}{300} + R \ln \frac{8}{2} \right) \]

### Step 9: Substitute Values
Using \(R = 8.314 \, \text{J/(mol K)}\):
\[ \Delta S = n \left( \frac{3}{2} \times 8.314 \ln 2 + 8.314 \ln 4 \right) \]

### Step 10: Calculate the Final Result
Using the provided logarithm values:
\[ \Delta S = n \left( \frac{3}{2} \times 8.314 \times 0.3 + 8.314 \times 1.386 \right) \]
Calculating the values:
\[ \Delta S \approx n \left( 3.745 + 11.528 \right) \]
\[ \Delta S \approx n \times 15.273 \]

Assuming 1 mole of gas:
\[ \Delta S \approx 15.273 \, \text{J/K} \]

### Final Answer
The entropy change (ΔS) in the process is approximately **15 J/K**.

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