Ideal Gases: Reversible Isothermal Process

Consider system with an ideal frictionless piston to apply and reduce pressure and change the volume. Recall that for an isothermal process that is reversible, the $P_{gas}=P_{ext}$. In this case the work in = heat out. Also, there is no net change in both $\Delta{U}$ and $\Delta{H}$, because the process occurs along the same isotherm, so $\Delta{T}=0$, as $T_1=T_2$.

 Therefore (a bit redundant, but we'll show the equations anyway):
  $$\Delta{U}=C_v(T_2-T_1)=0$$ $$\Delta{H}=C_p(T_2-T_1)=0$$

 Recall, the equation of work of an ideal gas $$w=RTln(V_1/V_2)$$

 Since the heat out is equal to the work in, the heat is given by $$q=-w=-RTln(V_1/V_2)=RTln(V_2/V_1)$$

 Note about the Concentration:

 When the number of moles are not equal to one, we cannot ignore the $n$ term. So when have some $P_1=n_1/V_1RT=C_1RT$ and $P_2=n_2/V_2RT=C_2RT$ where $C$ is the concentration.

Therefore, the ideal gas equation for work can also be expressed by $$w=nRTln(V_1/V_2)=NRTln(P_2/P_1)\text{ as } P_2/P_1=V_1/V_2$$ and $$w=nRTln(C_2/C_1)$$

 These equations come directly from the definitions of the terms. If they seem a bit confusing, refer to earlier physical chemistry posts on this blog, which can be found via the table of contents. Or if not, check wikipedia or something. :)