Question

During the derivation for work done in isothermal reversible expansion of an ideal gas, following expression appears.
$dW = \left( {P - dP} \right)dV = PdV\, + dP.dV$
A. True
B. False

Answer 1

Hint: The expression of work done in an isothermal reversible expansion of an ideal gas gives the work done during the change from initial to a final volume at a constant temperature.

Complete step by step answer:
Suppose an ideal gas is taken in a cylinder having a moving piston. When a expands at a constant temperature the volume of the gas increases. The piston moves upward. The gas does some work for this expansion.
The formula of work is as follows:
$W = - PdV$
Where,
W is the work done.
P is pressure.
dV is the change in volume.
The ideal gas equation is as follows;
$PV = \,nRT$
Where,
n is the number of molecules of gas.
R is gas constant.
T is the temperature.
Rearrange the ideal gas equation for pressure as follows:
$P\, = \,\dfrac{{nRT}}{V}$
Substitute the value of pressure in the work formula as follows:
$W = - \,\dfrac{{nRT}}{V}dV$
To determine the total change integrate the equation from initial volume ${V_i}$ to final volume.
$W = - \int\limits_{{V_I}}^{{V_f}} {\,\dfrac{{nRT}}{V}dV} $
The number of molecules for gas is constant and gas constant is already a constant and for isothermal process temperature is also constant.
So, take n, R and T out from the integration because they are constant.
$W = - nRT\int\limits_{{V_I}}^{{V_f}} {\,\dfrac{{dV}}{V}} $…… $(1)$
The solution to this type of integration is shown as follows:
$\int\limits_a^b {\,\dfrac{{dx}}{x}} = \,\left[ {\ln x} \right]_a^b$
Solve the integration of equation (1) as follows:
$W = - nRT\left[ {\ln V} \right]_{{V_I}}^{{V_f}}$
$W = - nRT\left[ {\ln {V_f} - \ln {V_i}} \right]$
$W = - nRT\ln \left[ {\dfrac{{{V_f}}}{{{V_i}}}} \right]$……(2)
Multiply the equation $(2)$ with $2.303$ to convert the\[\ln \]into \[\log \].
$W = - 2.303nRT\log \left[ {\dfrac{{{V_f}}}{{{V_i}}}} \right]$……(3)
Equations (2) and (3) represent the formula to calculate the work done in an isothermal reversible expansion of an ideal gas.
The expression $dW = \left( {P - dP} \right)dV = PdV\, + dP.dV$ does not appear during the derivation for work done in an isothermal reversible expansion of an ideal gas.

So, the correct answer is “Option B”.

Note: According to ideal gas equation, at constant temperature the ratio between volume and pressure is as follows:
$\dfrac{{{V_2}}}{{{V_1}}}\, = \dfrac{{{P_1}}}{{{P_2}}}$
So, $\dfrac{{{V_f}}}{{{V_i}}}$ can be replaced with $\dfrac{{{P_i}}}{{{P_f}}}$so, the expression of work done will be,
$W = - 2.303nRT\log \left[ {\dfrac{{{P_i}}}{{{P_f}}}} \right]$.