Consider n mole of an ideal gas contained in an insulating cylinder fitted with a perfectly insulating piston of area of cross section area A.
Let the volume of the gas be V and it exerts a pressure P on the walls of the cylinder and the piston . Due to this pressure ( if greater than the external pressure ) , the gas expands and piston moves through a small distance dx.
Force acting on the piston , F = PA
External work done by the gas is given by
dW = Fdx = PA dx = PdV
If the gas expands from initial volume ($V_{1}$) to the final volume ($V_{2}$), the total work
Step 1: Use the Adiabatic Condition
For a reversible adiabatic process:
$PV^\gamma = \text{constant} = K$
Also, from the ideal gas law:
$PV = nRT$
Step 2: Express Work as an Integral
The work done by the gas is:
$W = \int_{V_1}^{V_2} P \, dV$
Using the adiabatic condition ( $P = \frac{K}{V^\gamma}$), we get:
$W = \int_{V_1}^{V_2} \frac{K}{V^\gamma} \, dV = K \int_{V_1}^{V_2} V^{-\gamma} \, dV$
$W = K \left[ \frac{V^{1 - \gamma}}{1 - \gamma} \right]_{V_1}^{V_2} = \frac{K}{1 - \gamma} \left( V_2^{1 - \gamma} - V_1^{1 - \gamma} \right)$
Step 3: Replace \( K \) Using Initial Conditions
Since ( $K = P_1V_1^\gamma = P_2V_2^\gamma$), we substitute:
$W = \frac{P_2V_2^\gamma}{1 - \gamma} \left( V_2^{1 - \gamma} - V_1^{1 - \gamma} \right)$
Simplifying:
$W = \frac{1}{1 - \gamma} (P_2V_2 - P_1V_1)$
Or, more commonly:
$\boxed{W = \frac{P_1V_1 - P_2V_2}{\gamma - 1}}$
Work in terms of pressure and volume:
$W = \frac{P_1V_1 - P_2V_2}{\gamma - 1}$
Work in terms of temperature:
$W = \frac{nR(T_1 - T_2)}{\gamma - 1}$
Work in integral form:
$W = \int_{V_1}^{V_2} P \, dV = \frac{K}{1 - \gamma} \left( V_2^{1 - \gamma} - V_1^{1 - \gamma} \right)$