Question

Calculate the work done in kJ in a reaction, if volume of the reactant decreases from 8 dm\(^3\) to 4 dm\(^3\) against 43 bar pressure. [ 1 dm\(^3\) bar = 100J ]

Answer 1

Step 1: Recall the formula for work done.

Work done at constant pressure is: \[ w = -P \Delta V \]

Step 2: Calculate the change in volume.

Initial volume = \(V_i = 8 \, \text{dm}^3\) Final volume = \(V_f = 4 \, \text{dm}^3\) \[ \Delta V = V_f - V_i = 4 - 8 = -4 \, \text{dm}^3 \]

Step 3: Substitute values.

Given pressure: \(P = 43 \, \text{bar}\) \[ w = - (43) \times (-4) \, \text{dm}^3 \cdot \text{bar} \] \[ w = 172 \, \text{dm}^3 \cdot \text{bar} \]

Step 4: Convert units.

Since \(1 \, \text{dm}^3 \cdot \text{bar} = 100 \, J\): \[ w = 172 \times 100 = 17200 \, J \] \[ w = 17.2 \, \text{kJ} \]

Final Answer:

The work done is \[ \boxed{17.2 \, \text{kJ}} \]