Ideal Gases: Reversible Adiabatic Process

For a reversible adiabatic process, the system is moving between two different isotherms, so the temperature is changing. Also, the both volume and pressure are changing. So $P_1V_1T_1 \rightarrow P_2V_2T_2$.

The work done in an adiabatic process is less than that of an isothermal process, as an isothermal process is the maximum work possible to be done on the system. $w_{adiabatic} < w_{isothermal}$

For a reversible adiabatic process, there is no heat transfer to the system by definition, thus $q=0$. Since $\Delta{U}=q+w$, then $\Delta{U}=w=-PdV$.

For the following derivation, to make our lives a bit easier, we are going to consider the case where $n=1$. $$dU=-PdV$$ $$C_vdT=-PdV$$ $$C_vdT+PdV=0$$ $$C_vdT+RTdV/V=0$$ $$\text{as } P=RT/V$$ $$\text{divide equation by } T$$ $$C_vdT/T+RT/TdV/V=0/T$$ $$C_vdT/T+RdV/V=0$$ $$C_v\int_{T_1}^{T_2}dT/T+R\int_{V_1}^{V_2}dV/V=0$$ $$C_vln(T_2/T_1)+Rln(V_2/V_1)=0$$ $$\text{recall that }R=C_p-C_v$$ $$\text{divide by }C_v$$ $$C_v/C_vln(T_2/T_1)+(C_p-C_v)/C_vln(V_2/V_1)=0$$ $$ln(T_2/T_1)+(C_p/C_v+1)ln(V_2/V_1)=0$$ $$\text{let } C_p/C_v =\gamma$$ $$ln(T_2/T_1)+ln(V_2/V_1)^{\gamma-1}=0$$ $$ln(T_2/T_1)=ln(V_1/V_2)^{\gamma-1}$$ $$T_2/T_1=(V_1/V_2)^{\gamma-1}$$ $$\text{since } PV=RT$$ $$(P_2V_2)/(P_1V_1)=(V_1/V_2)^{\gamma-1}$$ $$P_2/P_1=(V_1/V_2)^{\gamma-1}(V_1/V_2)$$ $$P_2/P_1=(V_1/V_2)^{\gamma}$$ $$P_1V_1^{\gamma}=P_2V_2^{\gamma} \text{ where } \gamma = C_p/C_v$$  

To check our work, we can see if this applies to the ideal gas law. $$P_1V_1^{\gamma}=P_2V_2^{\gamma}$$ $$PV^{\gamma}=RTV^{\gamma}/V$$ $$PV^{\gamma}=RTV^{\gamma-1}$$ $$PV^{\gamma}/V^{\gamma}=RTV^{\gamma-1}/V^{\gamma}$$ $$P=RT/V$$ The gamma term can also be related to the number of degrees of freedom,$f$, in a gas. $$\gamma = C_p/C_v = \Delta{H}/\Delta{U}=(f+2)R/fR= (f+2)/f$$

To check out $\Delta{H}$, the entropy, we consider $$\Delta{H}=\Delta{U}+\Delta{PV}=w+q+\Delta{PV}=w+R\Delta{T}$$ $$\Delta{H}=C_v\Delta{T}+R\Delta{T}$$ $$\Delta{H}=(C_v+R)\Delta{T}$$ $$\Delta{H}=C_p\Delta{T}$$

In summary:
$$q=0$$ $$\Delta{U}=w=-PdV$$ $$\Delta{H}=C_p\Delta{T}$$ $$P_1V_1^{\gamma}=P_2V_2^{\gamma} \text{ where } \gamma = C_p/C_v$$