Air of initial volume V_0=0.1 m³ in a cylinder is under

Air of initial volume $V_0=0.1\ \text{m}^3$ in a cylinder is under pressure $P_0=10330\ \text{kgf/m}^2$ from a piston. Determine the work of isothermal compression of the air to volume $V_1=0.03\ \text{m}^3$.

Step 1: Pressure as a function of volume

Isothermal: $PV=P_0V_0$, so $P(V)=\dfrac{P_0V_0}{V}$.

Step 2: Work done by the gas during the volume change

\[W_{\text{by gas}}=\int_{V_0}^{V_1}P\,dV=P_0V_0\int_{V_0}^{V_1}\frac{dV}{V}=P_0V_0\ln\frac{V_1}{V_0}.\]

Step 3: Work done on the gas (compression)

Since $V_1 \lt V_0$ this is negative; the work required to compress the gas is its negative:

\[W=-W_{\text{by gas}}=P_0V_0\ln\frac{V_0}{V_1}.\]

\[\boxed{W=P_0V_0\ln\frac{V_0}{V_1}}\]

Step 4: Convert pressure to SI units

Treating $P_0=10330\ \text{kgf/m}^2$ as a force-per-area in kilogram-force (close to standard atmospheric pressure):

\[P_0=10330\times9.8\approx1.012\times10^5\ \text{Pa}.\]

Step 5: Numerical value

\[W=(1.012\times10^5)(0.1)\ln\frac{0.1}{0.03}=10123\times\ln(3.33)\approx10123\times1.204\approx12{,}190\ \text{J}\approx12.2\ \text{kJ}.\]

Step 6: Thermodynamic remark

For an isothermal process on an ideal gas, internal energy is unchanged ($\Delta U=0$), so by the first law all the work done on the gas is expelled as heat, $Q=-W\approx-12.2$ kJ. The logarithmic (rather than linear) dependence on the volume ratio reflects that pressure — and hence the force needed — rises as the gas is squeezed further.