JEE Main & Advanced Chemistry Thermodynamics / रासायनिक उष्मागतिकी Expansion Of An Ideal Gas

(1) Isothermal Expansion : For an isothermal expansion, \[\Delta T=0\]; \[\Delta E=\]0.

According to first law of thermodynamics,

\[\Delta E=q+w\]    \[\therefore \,q=-w\]

This shows that in isothermal expansion, the work is done by the system at the expense of heat absorbed.

Since for isothermal process, \[\Delta E\] and \[\Delta T\] are zero respectively, hence, \[\Delta H=0\]

(i) Work done in reversible isothermal expansion :  Consider an ideal gas enclosed in a cylinder fitted with a weightless and frictionless piston. The cylinder is not insulated. The external pressure, \[{{P}_{ext}}\] is equal to pressure of the gas, \[{{P}_{gas}}\].

\[{{P}_{ext}}={{P}_{gas}}=P\]

If the external pressure is decreased by an infinitesimal amount dP, the gas will expand by an infinitesimal volume, dV. As a result of expansion, the pressure of the gas within the cylinder falls to \[{{P}_{gas}}-dP\], i.e., it becomes again equal to the external pressure and, thus, the piston comes to rest. Such a process is repeated for a number of times, i.e., in each step the gas expands by a volume dV. 

Since the system is in thermal equilibrium with the surroundings, the infinitesimally small cooling produced due to expansion is balanced by the absorption of heat from the surroundings and the temperature remains constant throughout the expansion.

The work done by the gas in each step of expansion can be given as,      \[{{d}_{w}}=-({{P}_{ext}}-dP)dV=-{{P}_{ext}}.dV-dP\ .\ dV\]

\[dP.dV,\] the product of two infinitesimal quantities, is negligible.

The total amount of work done by the isothermal reversible expansion of the ideal gas from volume \[{{V}_{1}}\] to volume \[{{V}_{2}}\] is, given as, \[w=-nRT{{\log }_{e}}\frac{{{V}_{2}}}{{{V}_{1}}}\] or \[w=-2.303nRT{{\log }_{10}}\frac{{{V}_{2}}}{{{V}_{1}}}\]          

At constant temperature, according to Boyle’s law,

\[{{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}\] or \[\frac{{{V}_{2}}}{{{V}_{1}}}=\frac{{{P}_{1}}}{{{P}_{2}}}\] So,  \[w=-2.303nRT{{\log }_{10}}\frac{{{P}_{1}}}{{{P}_{2}}}\]           

Isothermal compression work of an ideal gas may be derived similarly and it has exactly the same value with positive sign.

\[{{w}_{compression}}=2.303nRT\log \frac{{{V}_{1}}}{{{V}_{2}}}=2.303nRT\log \frac{{{P}_{2}}}{{{P}_{1}}}\]

(ii) Work done in irreversible isothermal expansion : Two types of irreversible isothermal expansions are observed, i.e., (a) Free expansion and (b) Intermediate expansion. In free expansion, the external pressure is zero, i.e., work done is zero when gas expands in vacuum. In intermediate expansion, the external pressure is less than gas pressure. So, the work done when volume changes from \[{{V}_{1}}\] to \[{{V}_{2}}\] is given by

\[w=-\int_{{{V}_{1}}}^{{{V}_{2}}}{{{P}_{ext}}\times dV}=-{{P}_{ext}}({{V}_{2}}-{{V}_{1}})\]

Since \[{{P}_{ext}}\] is less than the pressure of the gas, the work done during intermediate expansion is numerically less than the work done during reversible isothermal expansion in which \[{{P}_{ext}}\] is almost equal to \[{{P}_{gas}}\].

(2) Adiabatic Expansion : In adiabatic expansion, no heat is allowed to enter or leave the system, hence, \[q=0\].

According to first law of thermodynamics,

\[\Delta E=q+w\]   \[\therefore \,\ \ \ \Delta E=w\]

work is done by the gas during expansion at the expense of internal energy. In expansion, \[\Delta E\] decreases while in compression \[\Delta E\] increases.

The molar specific heat capacity at constant volume of an ideal gas is given by

\[{{C}_{v}}={{\left( \frac{dE}{dT} \right)}_{v}}\]or \[dE={{C}_{v}}.dT\]

and for finite change \[\Delta E={{C}_{v}}\Delta T\] So,   \[w=\Delta E={{C}_{v}}\Delta T\] 

The value of \[\Delta T\] depends upon the process whether it is reversible or irreversible.

(i) Reversible adiabatic expansion : The following relationships are followed by an ideal gas under reversible adiabatic expansion.

\[P{{V}^{\gamma }}=\text{constant}\]

where, P = External pressure, V = Volume

\[\gamma =\frac{{{C}_{p}}}{{{C}_{v}}}\] 

where, \[{{C}_{p}}=\] molar specific heat capacity at constant pressure, \[{{C}_{v}}=\] molar specific heat capacity at constant volume.

\[{{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{\gamma }}={{\left( \frac{{{P}_{1}}}{{{P}_{2}}} \right)}^{\gamma -1}}={{\left( \frac{{{P}_{2}}}{{{P}_{1}}} \right)}^{1-\gamma }}\]

knowing \[\gamma \], \[{{P}_{1}},\,\,{{P}_{2}}\] and initial temperature \[{{T}_{1}}\], the final temperature \[{{T}_{2}}\] can be evaluated.

(ii) Irreversible adiabatic expansion : In free expansion, the external pressure is zero, i.e, work done is zero. Accordingly, \[\Delta E\] which is equal to w is also zero. If \[\Delta E\] is zero, \[\Delta T\] should be zero. Thus, in free expansion (adiabatically), \[\Delta T=0\], \[\Delta E=0\], \[w=0\] and \[\Delta H=0\].

In intermediate expansion, the volume changes from \[{{V}_{1}}\] to \[{{V}_{2}}\] against external pressure, \[{{P}_{ext}}\].

\[w=-{{P}_{ext}}({{V}_{2}}-{{V}_{1}})\]\[=-{{P}_{ext}}\left( \frac{R{{T}_{2}}}{{{P}_{2}}}-\frac{R{{T}_{1}}}{{{P}_{1}}} \right)\]

\[=-{{P}_{ext}}\left( \frac{{{T}_{2}}{{P}_{1}}-{{T}_{1}}{{P}_{2}}}{{{P}_{1}}{{P}_{2}}} \right)\times R\]

or \[w={{C}_{v}}({{T}_{2}}-{{T}_{1}})=-R{{P}_{ext}}\left( \frac{{{T}_{2}}{{P}_{1}}-{{T}_{1}}{{P}_{2}}}{{{P}_{1}}{{P}_{2}}} \right)\]