Name: Gibbs Free Energy, EMF and Equilibrium Constant
Uploaded: 2019-07-19
Duration: 6 min 15 s
Description: There are various thermodynamic parameters that are to be considered while studying the working of an electrochemical cell. Let's learn about them here!

Question

Derive the equation

W_{rev}

for isothermal reversible process.

Accepted Answer

Let us consider $n$ moles of ideal gas enclosed in a cylinder fitted with a weightless and frictionless piston. The work of expansion for a small change of volume $dV$ against the external pressure $P$ is given by-

$dW = - PdV$

$∴$ Total work done when the gas expands from initial volume $V_{1}$ to the final volume $V_{2}$ , will be-

$W = - \int_{V_{1}}^{V_{2}} PdV$

For an ideal gas,

$PV = nRT$

$\Rightarrow P = \frac{nRT}{V}$

Therefore,

$W = - \int_{V_{1}}^{V_{2}} \frac{nRT}{V} dV$

For isothermal expansion, $T$ is constant. Therefore,

$W = - nRT \int_{V_{1}}^{V_{2}} \frac{dV}{V}$

$\Rightarrow W = - nRT [lnV]_{v_{1}}^{v_{2}}$

$\Rightarrow W = - nRT ({lnV}_{2} - {lnV}_{1})$

$\Rightarrow W = - nRTln (\frac{V_{2}}{V_{1}})$

$\Rightarrow W = - 2.303 nRTlog (\frac{V_{2}}{V_{1}}) \dots . (1)$

At constant temperature,

$P_{1} V_{1} = P_{2} V_{2}$

$\Rightarrow \frac{V_{2}}{V_{1}} = \frac{P_{1}}{P_{2}}$

Therefore,

$W = - 2.303 nRTlog (\frac{P_{1}}{P_{2}}) \dots . (2)$

Equation $(1) & (2)$ are the expression for the work obtained in an isothermal reversible expansion of an ideal gas.

Derive the equation $W_{rev}$ for isothermal reversible process.

Important Questions on Thermodynamics

Learn and Prepare for any exam you want !

Derive the equation Wrev for isothermal reversible process.

Important Questions on Thermodynamics

Learn and Prepare for any exam you want !

Derive the equation $W_{rev}$ for isothermal reversible process.