To solve the problem, we need to calculate the work done during the adiabatic expansion of 1 mole of CO₂ gas that expands to 27 times its original volume at a temperature of 300 K. We will use the given values of \( \gamma = 1.33 \) and \( C_v = 25.10 \, \text{J mol}^{-1} \text{K}^{-1} \).
### Step-by-Step Solution:
1. **Identify Initial Conditions:**
- Number of moles, \( n = 1 \, \text{mol} \)
- Initial temperature, \( T_1 = 300 \, \text{K} \)
- Initial volume, \( V_1 = V \)
- Final volume, \( V_2 = 27V \)
2. **Use the Adiabatic Condition:**
For an adiabatic process, the relationship between temperature and volume is given by:
\[
\frac{T_1}{T_2} = \left(\frac{V_2}{V_1}\right)^{\gamma - 1}
\]
Substituting the known values:
\[
\frac{T_1}{T_2} = \left(27\right)^{1.33 - 1}
\]
\[
\frac{T_1}{T_2} = 27^{0.33}
\]
3. **Calculate \( 27^{0.33} \):**
Using a calculator, we find:
\[
27^{0.33} \approx 3
\]
Therefore,
\[
\frac{300}{T_2} = 3
\]
This implies:
\[
T_2 = \frac{300}{3} = 100 \, \text{K}
\]
4. **Calculate Change in Internal Energy (\( \Delta U \)):**
The change in internal energy for an ideal gas is given by:
\[
\Delta U = n C_v (T_2 - T_1)
\]
Substituting the values:
\[
\Delta U = 1 \times 25.10 \, \text{J mol}^{-1} \text{K}^{-1} \times (100 - 300)
\]
\[
\Delta U = 25.10 \times (-200) = -5020 \, \text{J}
\]
5. **Calculate Work Done (\( W \)):**
For an adiabatic process, the first law of thermodynamics states:
\[
\Delta U = Q + W
\]
Since it is an adiabatic process, \( Q = 0 \), thus:
\[
W = \Delta U
\]
Therefore, the work done is:
\[
W = -5020 \, \text{J}
\]
(The negative sign indicates work done by the system.)
6. **Convert Work to Kilojoules:**
To convert from joules to kilojoules:
\[
W = \frac{-5020}{1000} = -5.020 \, \text{kJ}
\]
The magnitude of work done is:
\[
|W| = 5.020 \, \text{kJ}
\]
7. **Round to the Nearest Whole Number:**
Rounding \( 5.020 \) to the nearest whole number gives:
\[
W \approx 5 \, \text{kJ/mol}
\]
### Final Answer:
The magnitude of work done during the expansion is **5 kJ/mol**.
