Question

Calculate q, w, \[\Delta U\] and \[\Delta H\] for the reversible isothermal expansion of one mole of an ideal gas at \[{{127}^{o}}C\] from a volume of \[10dc{{m}^{3}}\] to \[20dc{{m}^{3}}\].

Answer 1

Hint: Isothermal process is a process in which the temperature is kept constant. It is a thermodynamic process in which transfer of heat in and out of the system happens very slowly that a thermal equilibrium is maintained. So, for the isothermal process \[\Delta U=\Delta H=0\], here $\Delta U$ is the internal energy and $\Delta H$ is the change in the enthalpy.

Complete step-by-step answer:
For isothermal processes, we know that the change in internal energy and the change in enthalpy are equal to zero.
\[\Delta U=\Delta H=0\]
According to first law of thermodynamics,
\[\Delta U=q+w=0\]
\[q=-w\]
We know that the equation for work done by the system in relation with the volume of the gas is given by
\[w=-2.303nRT\log \dfrac{{{V}_{2}}}{{{V}_{1}}}\]
Where, \[{{V}_{2}}\]and \[{{V}_{1}}\] are the initial and final volumes, n is the number of moles involved, R is Rydberg’s constant and T is the temperature \[{{127}^{o}}C\]=127+273=400K
\[{{V}_{1}}\]=\[10dc{{m}^{3}}\]\[{{V}_{2}}\]=\[20dc{{m}^{3}}\]
Substituting the values in the equation we get
\[w=-2.303\times 1\times 8.314\times 400\times \log \dfrac{20}{10}\]
=\[-2305.3J\]
This is the work done by the system.
Now, the heat absorbed by the system is q.
\[q=-w\]
\[q=2305.3J\]
Hence, \[q=2305.3J\], w=\[-2305.3J\] and \[\Delta U=\Delta H=0\].

Additional Information: In an isothermal process \[\Delta T=0\]. Here are some examples of isothermal processes. It occurs in almost any kind of system which has a means of regulating temperature, like in machines and in living cells, etc. Some cycles of the heat engine are isothermal. In an isothermal expansion, the energy given to the system does work on the surrounding. In case of isothermal compression, the work is done on the system to decrease the volume and increase the pressure.

Note: As temperature is kept constant in the isothermal process, there is no heat exchange. So, there is no enthalpy change and internal energy occurring in the process. Both will be always zero in an isothermal process.