. A piston has an external pressure of 12.0 atm. How much work has been done in joules if the cylinder goes from a volume of 0.120 liters to 0.650 liters? Express your answer with the appropriate units.
The Correct Answer and Explanation is:
Work done w=−Pext ΔV=−12.0 atm (0.650 L−0.120 L)=−12.0 atm (0.530 L)=−6.36 L\cdotpatmw = -P_{\text{ext}}\;\Delta V = -12.0\ \text{atm}\;(0.650\ \text{L} – 0.120\ \text{L}) = -12.0\ \text{atm}\;(0.530\ \text{L}) = -6.36\ \text{L·atm}
Converting L\cdotpatm\text{L·atm} to joules ( 1 L\cdotpatm=101.325 J1\ \text{L·atm} = 101.325\ \text{J} ): w=−6.36 L\cdotpatm (101.325 J L−1atm−1)≈−6.44×102 Jw = -6.36\ \text{L·atm}\;(101.325\ \text{J L}^{-1}\text{atm}^{-1}) \approx -6.44 \times 10^{2}\ \text{J} w≈−6.44×102 J\boxed{w \approx -6.44 \times 10^{2}\ \text{J}}
Why this result makes sense
In thermodynamics, the mechanical work associated with a gas pushing back an external force is defined (in chemistry) as
w=−PextΔVw = -P_{\text{ext}}\Delta V.
- Sign convention. The negative sign reflects the “system” point of view adopted in chemistry: if the gas expands, it spends energy doing work on its surroundings, and the system’s internal energy decreases, so ww is negative. If the gas were compressed (ΔV<0\Delta V < 0), ww would be positive because energy is supplied to the gas.
- Constant external pressure. Because the problem states a single external pressure (12.0 atm), we do not need to integrate P(V)P(V) over the entire path. The work is simply the product of that constant pressure and the change in volume.
- Volume change. The cylinder grows from 0.120 L to 0.650 L, so ΔV=+0.530 L\Delta V = +0.530\ \text{L}. A positive ΔV\Delta V confirms this is an expansion.
- Unit conversion. Pressure–volume work is naturally expressed in L·atm, but energy must be in joules to match SI. The experimentally determined equivalence 1 L\cdotpatm=101.325 J1\ \text{L·atm} = 101.325\ \text{J} comes directly from the definition of the joule and the pascal (1 Pa = 1 N m−2^{-2}). By multiplying by this conversion factor, we translate macroscopic mechanical work into the SI energy unit.
- Magnitude. 644 J is roughly the energy required to lift a 6.6 kg mass one metre against Earth’s gravity (because E=mgh≈6.6 kg×9.81 m s−2×1 m≈65 JE = mgh \approx 6.6\ \text{kg}\times9.81\ \text{m s}^{-2}\times1\ \text{m} ≈ 65\ \text{J} per kilogram‐metre). This comparison helps to visualize that high pressures acting over even modest laboratory volumes can involve hundreds of joules—substantial on a human scale.
- Significant figures. Each measured quantity (12.0 atm, 0.120 L, 0.650 L) has three significant figures, so the final answer is reported to three significant figures: 6.44×102 J6.44 \times 10^{2}\ \text{J}.
Thus, the expanding gas has done about −6.44 × 10² J of work on its surroundings.
