0.5 mole of `SO_2` gas at `27^@C` is expanded in reversible adiabatic condition to make volume 8 times.Final temperature (in K) and work done by gas (in Cal) respectively are x and y.Report your answer as `y/x`.
(Take ideal behaviour of gas, R=2 Cal/K mol and `2^(1.2)=2.3`)

Explanation

To solve the problem, we need to find the final temperature (T2) and the work done (W) during the reversible adiabatic expansion of 0.5 moles of SO2 gas. We will then report the ratio of work done to final temperature (Y/X).

### Step 1: Convert Initial Temperature to Kelvin
The initial temperature (T1) is given as 27°C. To convert this to Kelvin:
\[ T_1 = 27 + 273 = 300 \, K \]

**Hint:** Remember to always convert Celsius to Kelvin by adding 273.

### Step 2: Determine the Volume Change
The volume is expanded to 8 times its original volume:
\[ V_2 = 8V_1 \]

### Step 3: Determine the Value of Gamma (γ)
For SO2, the value of gamma (γ) is given as 1.33.

### Step 4: Use the Adiabatic Relation
The adiabatic condition for an ideal gas can be expressed as:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
Substituting the known values:
\[ 300 \cdot V^{1.33 - 1} = T_2 \cdot (8V)^{1.33 - 1} \]
This simplifies to:
\[ 300 \cdot V^{0.33} = T_2 \cdot 8^{0.33} \cdot V^{0.33} \]
Cancelling \(V^{0.33}\) from both sides:
\[ 300 = T_2 \cdot 8^{0.33} \]

### Step 5: Calculate \(8^{0.33}\)
Using the provided approximation \(2^{1.2} = 2.3\):
\[ 8^{0.33} = (2^3)^{0.33} = 2^{1} = 2 \]

### Step 6: Solve for T2
Now substituting back:
\[ 300 = T_2 \cdot 2 \]
\[ T_2 = \frac{300}{2} = 150 \, K \]

### Step 7: Calculate Work Done (W)
The work done in an adiabatic process is given by:
\[ W = nC_v (T_2 - T_1) \]
Where:
- \(n = 0.5 \, \text{moles}\)
- \(C_v\) for SO2 is given as \(3R\). Since \(R = 2 \, \text{Cal/K mol}\), we have:
\[ C_v = 3 \times 2 = 6 \, \text{Cal/K mol} \]

Now substituting the values:
\[ W = 0.5 \cdot 6 \cdot (150 - 300) \]
\[ W = 0.5 \cdot 6 \cdot (-150) \]
\[ W = -450 \, \text{Cal} \]

### Step 8: Calculate the Ratio \(Y/X\)
Here, \(Y\) is the work done and \(X\) is the final temperature:
\[ Y = -450, \, X = 150 \]
Thus,
\[ \frac{Y}{X} = \frac{-450}{150} = -3 \]

### Final Answer
The final answer is:
\[ \frac{Y}{X} = -3 \]

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