Question

Derive the equation for the work of expansion for an isothermal reversible process.

Answer 1

The "isothermal process", which is thermodynamic process in which the temperature of a system remains constant.

Let us consider a case when we are given n moles of gas and thermal equilibrium is maintained

so by ideal gas equation we can write $pV=nRT$

so work done is given by $dW=-pdV$

$p$ can be substituted as $nRT/V$

then $dW$ can be written as

$dW=-(nRT/V)dV$

$\intop dW=-\intop ^{V_f}_{V_i}(nRT/V)dV$

$W=-nRT[ln V]^{V_f}_{V_i}$

$W=-nRT(lnV_f-lnV_i)$

$W=-2.303*nRT(logV_f/logV_i)$