Question

A gas follows a general process as \[PV = RT + 3V\] for 1 mole gas. If it expands
isobarically till temperature is doubled, then the work done by the gas is (initial temperature and pressure are \[{T_0}\] and \[{P_0}\] respectively)
A. \[\dfrac{{{P_0}{T_0}R}}{{\left( {2{P_0} - 3} \right)}}\]
B. \[\dfrac{{{P_0}{T_0}R}}{{\left( {{P_0} - 3} \right)}}\]
C. \[\dfrac{{{P_0}{T_0}R}}{{\left( {{P_0}V - 3} \right)}}\]
D. \[\dfrac{{3{P_0}{V_0}}}{R}\]

Answer 1

Hint:Differentiate the given equation to determine the small change in the volume with respect to temperature change. The work done in the isobaric process is the integration of product of pressure and volume change. Substitute the limits of integration from \[{T_0}\] to
\[2{T_0}\] and calculate the work done for the given isobaric process.

Formula used:
Work done,\[W = \int {P\,dV} \]

Here, P is the pressure and dV is the small change in the volume.

Complete step by step answer:

We know that in an isobaric process, the pressure of the gas remains constant. We have given
the equation for the isobaric process,
\[PV = RT + 3V\]

Here, P is the pressure and V is the volume of the gas.

We can write the above equation for the initial state of the gas as follows,
\[{P_0}V = RT + 3V\]

Differentiating the above equation, we get,
\[{P_0}dV = RdT + 3dV\]
\[ \Rightarrow \left( {{P_0} - 3} \right)dV = RdT\]

\[ \Rightarrow dV = \dfrac{{R\,dT}}{{{P_0} - 3}}\]

We have the expression for the work done in isobaric process,
\[W = \int {P\,dV} \]
\[ \Rightarrow W = {P_0}\int {\dfrac{{R\,dT}}{{{P_0} - 3}}} \]
\[ \Rightarrow W = \dfrac{{{P_0}R}}{{{P_0} - 3}}\int {dT} \]

Substituting the limits of integration from \[{T_0}\] to \[2{T_0}\], we get,
\[W = \dfrac{{{P_0}R}}{{{P_0} - 3}}\int\limits_{{T_0}}^{2{T_0}} {dT} \]
\[ \Rightarrow W = \dfrac{{{P_0}R}}{{{P_0} - 3}}\left( T \right)_{{T_0}}^{2{T_0}}\]
\[ \Rightarrow W = \dfrac{{{P_0}R}}{{{P_0} - 3}}\left( {2{T_0} - {T_0}} \right)\]
\[ \Rightarrow W = \dfrac{{{P_0}{T_0}R}}{{\left( {{P_0} - 3} \right)}}\]

So, the correct answer is option (B).

Note: We have the ideal gas equation, \[PV = nRT\], where, n is the number of moles of the gas and R is the gas constant. The given gas does not follow this equation and is said to be non-ideal gas. The work done expression \[W = \int {P\,dV} \] is only for the process where the pressure of the gas does not change and only the volume or the temperature change.