Proof question

[wscha004]

So I have this problem in my Thermodynamics homework where you have to prove that one equation is equal to another by using different constants and equalities. Well, my professor posted the answer but there is a step that I am pretty sure uses algebra to rearrange the formula and I dont understant what has been done. The originial term is "R*ln*[(T2*P1)/(T1*P2)]". Then they jump to the next step and say that this is can be rearranged into "R*ln*(T2/T1) - R*ln*(P2/P1)" and I dont understand how this jump was made. Note: In the terms such as "P2, T1" etc, the numbers are subscripts to denote first temperature, second pressure and so on. Also the "R" in the equation is just a constant for an ideal gas. Thanks in advance!!!

 

[Loren]

ln*[(T2*P1)/(T1*P2)]=R*ln*(T2/T1) - R*ln*(P2/P1)?
Based on \(\displaystyle \ln(ab)=\ln a + \ln b\)
and
\(\displaystyle \ln(\frac{a}{b})=\ln a - \ln b\)

\(\displaystyle \ln(\frac{T_2P_1}{T_1P_2})=\ln(T_2P_1)-\ln(T_1P_2)=\ln T_2 + \ln P_1 - (\ln T_1 + \ln P_2)=\ln T_2 + \ln P_1 - \ln T_1 - \ln P_2 =\)
\(\displaystyle \ln T_2 - \ln T_1 + \ln P_1 - \ln P_2 = \ln(\frac{T_2}{T_1}) + \ln (\frac{P_1}{P_2})\)

 

[wscha004]

Thank you so much!!! I knew there was some algebra involved in that but I could not remember any of the natural log relationtions. Thanks again!!!