Question

Calculate the work (kJ) done during a reaction in which the internal volume expands from 20 L to 43 L against an outside pressure of 4.3 atm.

Answer 1

Answer:

$\large \boxed{\text{-10.0 kJ}}$

Explanation:

1. Calculate the work

w = - pΔV = -4.3 atm × (43 L - 20 L) = -4.3 × 23 L·atm = -98.9 L·atm

2. Convert litre-atmospheres to joules

$w = \text{-98.9 L\cdot atm } \times \dfrac{\text{101.3 J}}{\text{1 L \cdot atm }} = \text{-10000 J} = \textbf{-10.0 kJ}\\\\\text{The work done is \large \boxed{\textbf{-10.0 kJ}} }$

The negative sign indicates that the work was done against the surroundings.