Question

Work done in reversible isothermal process by an ideal gas is given by:

• A. 2.303 nRT $log\frac{V_2}{V_1}$
• B. $\frac{nR}{\left(\gamma -1\right)}(T_2-T_1)$
• C. 2.303 nRT $log\frac{V_1}{V_2}$
• D. none

Answer 1

Answer: (C)

2.303 nRT $log\frac{V_1}{V_2}$

Solution:- (C) $2.303nRT\times \log \left(\frac{V_1}{V_2}\right)$

Consider '$n$' moles of an ideal gas enclosed in a cylinder fitted with a weightless, frictionless, airtight movable position. Let the pressure of the gas be $P$ which is equal to external atmospheric pressure $P$. Let the external pressure be reduced by an infinitely small amount $dP$ and the corresponding small increase in volume be $dV$.

Therefore the small work done in the expnsion process will be-

$dW=-P_{ext.}dV$

$\Rightarrow dW=-(P-dP)dV$

$\Rightarrow dW=-P.dV+dP.dV$

Since both $dP$ and $dV$ are very small, the product $dP.dV$ will be very small in comparison with $P.dV$ and thus can be neglected.

$\therefore dW=-P.dV$

When the expansion of the gas is carried out reversibly then there will be a series of such $P.dV$ terms. Thus the total maximum work $W_{max}$ can be obtained by integrating this equation between the limits $V_1$ to $V_2$. Where $V_1$ is initial volume while $V_2$ is final value.

$W={\int }_{V_1}^{V_2}dW$

$W={\int }_{V_1}^{V_2}(-P.dV)$

As we know that, from ideal gas equation,

$PV=nRT$

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

$W=-{\int }_{V_1}^{V_2}\frac{nRT}{V}dV$

$W=-nRT{\int }_{V_1}^{V_2}\frac{dV}{V}$

$W=-nRT{\left[\ln V\right]}_{V_1}^{V_2}$

$W=-nRT(\ln V_2-\ln V_2)$

$W=nRT(\ln V_1-\ln V_2)$

$W=nRT\ln \left(\frac{V_1}{V_2}\right)$

$W=nRT(2.303\times \log \left(\frac{V_1}{V_2}\right))$

$W=2.303nRT\times \log \left(\frac{V_1}{V_2}\right)$

Hence work done in reversible isothermal process by an ideal gas is given by-

$W=2.303nRT\times \log \left(\frac{V_1}{V_2}\right)$