1 Mole of `CO_(2)` gas at 300 K expanded under that reversible adiabtic condition such that its volume becomes 27 times. The magnitude of work (in kJ/mol) is :
`("Given "y=1.33 and C_(v)="25.10 J mol"^(-1)K^(-1)" for "CO_(2))` report your answer by rounding it up to nearest whole number

Answer

To solve the problem, we need to calculate the work done during the reversible adiabatic expansion of 1 mole of CO₂ gas. Here’s a step-by-step solution:

### Step 1: Identify Given Values
- Number of moles (n) = 1 mole
- Initial temperature (T₁) = 300 K
- Final volume (V₂) = 27 × V₁
- Heat capacity at constant volume (Cᵥ) = 25.10 J/(mol·K)
- Heat capacity ratio (γ) = 1.33

### Step 2: Use the Adiabatic Condition
For an adiabatic process, the relationship between temperature and volume is given by:
\[ \frac{T₁}{T₂} = \left(\frac{V₂}{V₁}\right)^{\gamma - 1} \]
We can rearrange this to find T₂:
\[ T₂ = T₁ \left(\frac{V₁}{V₂}\right)^{\gamma - 1} \]

### Step 3: Substitute Values
Substituting the known values into the equation:
- \( V₂ = 27V₁ \)
- Therefore, \( \frac{V₁}{V₂} = \frac{1}{27} \)

Now substituting into the equation for T₂:
\[ T₂ = 300 \left(\frac{1}{27}\right)^{1.33 - 1} \]
Calculating \( \gamma - 1 \):
\[ \gamma - 1 = 1.33 - 1 = 0.33 \]
Thus,
\[ T₂ = 300 \left(\frac{1}{27}\right)^{0.33} \]

### Step 4: Calculate \( \left(\frac{1}{27}\right)^{0.33} \)
Calculating \( \left(27\right)^{-0.33} \):
\[ \left(27\right)^{-0.33} \approx \frac{1}{3} \quad \text{(since } 27^{1/3} = 3\text{)} \]
Therefore,
\[ T₂ \approx 300 \times \frac{1}{3} = 100 \text{ K} \]

### Step 5: Calculate Work Done
Using the first law of thermodynamics:
\[ \Delta E = Q + W \]
Since \( Q = 0 \) for an adiabatic process, we have:
\[ W = \Delta E = nCᵥ(T₂ - T₁) \]
Substituting the values:
\[ W = 1 \times 25.10 \, \text{J/(mol·K)} \times (100 - 300) \text{ K} \]
Calculating:
\[ W = 25.10 \times (-200) = -5020 \text{ J/mol} \]

### Step 6: Convert to kJ/mol
To convert Joules to kilojoules:
\[ W = -5020 \text{ J/mol} = -5.02 \text{ kJ/mol} \]

### Step 7: Report the Magnitude
The magnitude of work done is:
\[ |W| = 5.02 \text{ kJ/mol} \]
Rounding to the nearest whole number gives:
\[ \text{Magnitude of work} = 5 \text{ kJ/mol} \]

### Final Answer
The magnitude of work done during the expansion is **5 kJ/mol**.