The work done during the expansion of a gas from a volume of $4 ext{ dm}^3$ to $6 ext{ dm}^3$ against a cons

Question

The work done during the expansion of a gas from a volume of $4	ext{ dm}^3$ to $6	ext{ dm}^3$ against a constant external pressure of $3	ext{ atm}$ is:

Accepted Answer

The work done ($w$) during the expansion of a gas against a constant external pressure ($p_{ext}$) is given by the formula:
$w = -p_{ext} \Delta V = -p_{ext}(V_f - V_i)$
Given:
Initial volume, $V_i = 4	ext{ dm}^3 = 4	ext{ L}$ (since $1	ext{ dm}^3 = 1	ext{ L}$)
Final volume, $V_f = 6	ext{ dm}^3 = 6	ext{ L}$
External pressure, $p_{ext} = 3	ext{ atm}$
Substituting the values:
$w = -3	ext{ atm} 	imes (6	ext{ L} - 4	ext{ L}) = -3 	imes 2 = -6	ext{ L atm}$
To convert the work done into Joules, we use the conversion factor $1	ext{ L atm} = 101.325	ext{ J}$:
$w = -6	ext{ L atm} 	imes 101.325	ext{ J/L atm} = -607.95	ext{ J} \approx -608	ext{ J}$.

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