recall and use W = p∆V for the work done when the volume of a gas changes at constant pressure and understand the difference between the work done by the gas and the work done on the gas

First Law of Thermodynamics

The first law is a statement of energy conservation for a closed system. It tells us that any change in the internal energy of a system comes from heat added to the system or work done on it.

Work Done by a Gas at Constant Pressure

When a gas expands or compresses while the external pressure stays the same, the work done by the gas is simply the area under the pressure–volume curve, which is a rectangle. The formula is:

\$W = p\,\Delta V\$

• \$p\$ = external pressure (Pa)

• \$\Delta V = Vf - Vi\$ (m³)

If the gas expands (\$\Delta V>0\$), \$W\$ is positive: the gas does work on the surroundings.

If it compresses (\$\Delta V<0\$), \$W\$ is negative: work is done on the gas.

Work Done By the Gas vs Work Done On the Gas

SituationWork Done By Gas (W)Work Done On Gas (−W)
Expansion (\$\Delta V>0\$)\$+p\Delta V\$ (positive)\$-p\Delta V\$ (negative)
Compression (\$\Delta V<0\$)\$-p\Delta V\$ (negative)\$+p\Delta V\$ (positive)

Exam Tips 📚

  • Remember the sign convention: work done by the system is positive when the system expands.

  • When given a pressure and change in volume, plug into \$W = p\Delta V\$ directly.

  • Check units: \$p\$ in Pa, \$\Delta V\$ in m³, so \$W\$ is in joules.

  • For variable pressure, integrate \$W = \int p\,dV\$.

  • Use a piston diagram to visualise expansion vs compression.

Example Problem 🚀

A piston contains 2.0 L of gas at 1.0 bar. The piston is pushed in until the volume is 1.0 L. Calculate the work done on the gas.

Solution:

  1. Convert volumes to cubic metres: \$Vi = 2.0\,\text{L} = 2.0\times10^{-3}\,\text{m}^3\$, \$Vf = 1.0\times10^{-3}\,\text{m}^3\$.

  2. Pressure in pascals: \$p = 1.0\,\text{bar} = 1.0\times10^5\,\text{Pa}\$.

  3. Change in volume: \$\Delta V = Vf - Vi = -1.0\times10^{-3}\,\text{m}^3\$.

  4. Work: \$W = p\Delta V = 1.0\times10^5 \times (-1.0\times10^{-3}) = -100\,\text{J}\$.

  5. Interpretation: \$-100\$ J means 100 J of work is done on the gas (compression).

Analogy: Balloon & Air Pump 🎈

Imagine blowing up a balloon. The air inside does work on the rubber as it expands. If you squeeze the balloon, the rubber does work on the air (work done on the gas). The sign of the work changes with the direction of the volume change.

Key Takeaways

  • Work at constant pressure: \$W = p\Delta V\$.

  • Positive \$W\$ → gas expands, does work on surroundings.

  • Negative \$W\$ → gas compresses, work is done on gas.

  • Always keep track of units and sign conventions.