Question

Calculate the work (kJ) done during a reaction in which the internal volume expands from 30 L to 51 L against a vacuum (an outside pressure of 0 atm).

Answer 1

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**.