A cylinder has 40.0 cm radius and is 50.0 cm deep. It is filled with air at  and 1.00 atm (figure a). A 20.0 kg piston is now slowly lowered into the cylinder compressing the air trapped inside (figure b). Finally, a 75.0 kg man stands on the piston farther compressing the air, without any appreciable change in temperature. The atmospheric pressure  ${\mathrm{p}}_0=1\times {10}^5\mathrm{N}/{\mathrm{m}}^2$  Then, 

Since area of cross-section is constant, volume of the gas is proportional to height of the cylinder. When the gas is compressed by the piston, we have PV = constant, because temperature remains at 20^oC .
  $\therefore \mathrm{h}=\frac{0.5\times {10}^5}{1.00398\times {10}^5}=\frac{0.5}{1.004}=0.498\mathrm{m}\frac{{\mathrm{V}}_1}{{\mathrm{T}}_1}=\frac{{\mathrm{V}}_2}{{\mathrm{T}}_2}\text{ i.e., }\frac{0.498}{293}=\frac{0.498}{{\mathrm{T}}_2}$
the temperature must be raised to 24.2^oC When the gas is heated, the process is isobaric; when the piston compresses the gas, the process is isothermal.