# why isothermal expansion work done is more than the adibatic expansion work?

SAGAR SINGH - IIT DELHI
878 Points
13 years ago

Dear student,

In thermodynamics, the work involved when a gas changes from state A to state B is simply

$W_{A\to B} = \int_{V_A}^{V_B}p\,dV$

For an isothermal,reversible  process, this integral equals the area under the relevant pressure-volume isotherm, and is indicated in yellow in the figure (at the bottom right-hand of the page) for an ideal gas. Again, p = nRT / V applies and with T being constant (as this is an isothermal process), we have:

$W_{A\to B} = \int_{V_A}^{V_B}p\,dV = \int_{V_A}^{V_B}\frac{nRT}{V}dV = nRT\int_{V_A}^{V_B}\frac{1}{V}dV = nRT\ln{\frac{V_B}{V_A}}$

By convention, work is defined as the work the system does on its environment. If, for example, the system expands by a piston moving in the direction of force applied by the internal pressure of a gas, then the work is counted as positive, and as this work is done by using internal energy of the system, the result is that the internal energy decreases. Conversely, if the environment does work on the system so that its internal energy increases, the work is counted as negative.