Question

Work done in reversible isothermal process by an ideal gas is given by:
A.$2.303\,nRT\log \dfrac{{{V_2}}}{{{V_1}}}$
B.$\dfrac{{nR}}{{\left( {\gamma - 1} \right)}}\left( {{T_2} - {T_1}} \right)$
C.$2.303nRT\log \dfrac{{{V_1}}}{{{V_2}}}$
D.None

Answer 1

All the four properties of gas such as pressure, volume, temperature and number of moles are combined into a single equation is said to be Ideal gas law.
$PV = nRT$
Where
P indicates the pressure of the gas.
V indicates the volume of the gas.
n indicates the number of moles of the gas.
R indicates the universal gas constant.
T indicates the temperature in Kelvin.

Complete step by step answer:
Let us consider 'n' moles of an ideal gas surrounded in a cylinder fitted with a weightless, frictionless, airtight movable position.
Let us consider the pressure of the gas to be P that is equal to external atmospheric pressure P.
Let us consider the external pressure to be decreased by an infinitely small amount dP and the corresponding small increase in volume be dV.
Therefore the work done in the expansion process will be given as,
$dW = - {P_{ext}}dV$
On substituting the values we get,
$dW = - \left( {P - dP} \right)dV$
And
$dW = - P.dV + dP.dV$
Since both dP and dV are very small, the product dP.dV would be very small in comparison with P.dV and therefore, we can neglect it.
$dW = - P.dV$
When the expansion of the gas is carried out reversibly then there would be a series of such P.dV terms. Therefore, the total maximum work ${W_{\max }}$ could be obtained by integrating this equation between the limits ${V_1}$ and ${V_2}$.
The initial volume is represented as ${V_1}$ and the final volume is represented as ${V_2}$.
$W = \int_{{V_1}}^{{V_2}} {dW}$
On substituting the dW values we get,
$W = \int_{{V_1}}^{{V_2}} {\left( { - P.dV} \right)}$
From the ideal gas equation, we know that,
$PV = nRT$
Rearranging this equation, with respect to pressure, we get
$P = \dfrac{{nRT}}{V}$
Substituting the expression of P in the work done equation, we get
$W = \int_{V1}^{{V_2}} {\dfrac{{nRT}}{V}} dV$
When we integrate the equation, we get the equation as,
$W = - nRT\int_{{V_1}}^{{V_2}} {\dfrac{{dV}}{V}}$
We can write the above equation as,
$W = - nRT\left[ {\ln V} \right]_{{V_1}}^{{V_2}}$
$\Rightarrow W = - nRT\left( {\ln {V_2} - \ln {V_1}} \right)$
We can write the above equation as,
$W = nRT\left( {\ln {V_1} - \ln {V_2}} \right)$
On simplifying we get,
$W = nRT\ln \left( {\dfrac{{{V_1}}}{{{V_2}}}} \right)$
Convert ln to log using the value,
$W = nRT\left( {2.303 \times \log \left( {\dfrac{{{V_1}}}{{{V_2}}}} \right)} \right)$
$W = 2.303nRT \times \log \dfrac{{{V_1}}}{{{V_2}}}$
Therefore, the work done in reversible isothermal process using an ideal gas is written as,
$W = 2.303nRT \times \log \dfrac{{{V_1}}}{{{V_2}}}$
Hence Option (C) is correct.

Note:
We need to remember that the isothermal processes could take place in any type of system which contains some means of regulating the temperature, including highly structured machines, and living cells. Few parts of the cycles of certain heat engines are carried out isothermally. In an isothermal process, the internal energy of an ideal gas is fixed. In the isothermal compression of a gas there is work done on the system to reduce the volume and raise the pressure.