To calculate the work done during a reaction in which the internal volume expands, we can use the formula for work done during expansion against a constant external pressure:
\[ W = -P_{\text{ext}} \Delta V \]
Where:
- \( W \) is the work done.
- \( P_{\text{ext}} \) is the external pressure.
- \( \Delta V \) is the change in volume.
Given:
- The external pressure \( P_{\text{ext}} = 0 , \text{atm} \).
- The initial volume \( V_i = 30 , \text{L} \).
- The final volume \( V_f = 51 , \text{L} \).
First, we need to calculate the change in volume \( \Delta V \):
\[ \Delta V = V_f - V_i = 51 , \text{L} - 30 , \text{L} = 21 , \text{L} \]
Now we plug in the values into the work formula. Since the external pressure is 0 atm, the work done (W) is:
\[ W = -P_{\text{ext}} \Delta V = -0 , \text{atm} \cdot 21 , \text{L} = 0 \]
Since the work done is zero, the work done during this reaction where the volume expands against a vacuum is:
\[ W = 0 , \text{kJ} \]
Therefore, the work done during the reaction is 0 kJ.