--- title: "Adiabatic process" chunk: 5/7 source: "https://en.wikipedia.org/wiki/Adiabatic_process" category: "reference" tags: "science, encyclopedia" date_saved: "2026-05-05T10:56:55.166101+00:00" instance: "kb-cron" --- P = P 1 ( V 1 V ) γ . {\displaystyle P=P_{1}\ \left({\frac {V_{1}}{V}}\right)^{\gamma }~.} Substituting this into (b2) gives W = ∫ V 1 V 2 P 1 ( V 1 V ) γ d V . {\displaystyle W=\int _{V_{1}}^{V_{2}}P_{1}\ \left({\frac {V_{1}}{V}}\right)^{\gamma }\ \mathrm {d} V~.} Integrating, we obtain the expression for work, W = P 1 V 1 γ V 2 1 − γ − V 1 1 − γ 1 − γ = P 2 V 2 − P 1 V 1 1 − γ . {\displaystyle {\begin{aligned}W&=P_{1}\ V_{1}^{\gamma }\ {\frac {V_{2}^{1-\gamma }-V_{1}^{1-\gamma }}{1-\gamma }}\\[1ex]&={\frac {P_{2}\ V_{2}-P_{1}\ V_{1}}{1-\gamma }}.\end{aligned}}} Substituting γ = ⁠α + 1/α⁠ in the second term, W = − α P 1 V 1 γ ( V 2 1 − γ − V 1 1 − γ ) . {\displaystyle W=-\alpha \ P_{1}\ V_{1}^{\gamma }\ \left(V_{2}^{1-\gamma }-V_{1}^{1-\gamma }\right)~.} Rearranging, W = − α P 1 V 1 ( ( V 2 V 1 ) 1 − γ − 1 ) . {\displaystyle W=-\alpha \ P_{1}\ V_{1}\ \left(\left({\frac {V_{2}}{V_{1}}}\right)^{1-\gamma }-1\right)~.} Using the ideal gas law and assuming a constant molar quantity (as often happens in practical cases), W = − α n R T 1 ( ( V 2 V 1 ) 1 − γ − 1 ) . {\displaystyle W=-\alpha \ n\ R\ T_{1}\left(\left({\frac {V_{2}}{V_{1}}}\right)^{1-\gamma }-1\right)~.} By the continuous formula, P 2 P 1 = ( V 2 V 1 ) − γ , {\displaystyle {\frac {P_{2}}{P_{1}}}=\left({\frac {V_{2}}{V_{1}}}\right)^{-\gamma }\ ,} or ( P 2 P 1 ) − 1 γ = V 2 V 1 . {\displaystyle \left({\frac {P_{2}}{P_{1}}}\right)^{-{\frac {1}{\gamma }}}={\frac {V_{2}}{V_{1}}}~.} Substituting into the previous expression for W, W = − α n R T 1 ( ( P 2 P 1 ) γ − 1 γ − 1 ) . {\displaystyle W=-\alpha \ n\ R\ T_{1}\left(\left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}-1\right)~.} Substituting this expression and (b1) in (b3) gives α n R ( T 2 − T 1 ) = α n R T 1 ( ( P 2 P 1 ) γ − 1 γ − 1 ) . {\displaystyle \alpha \ n\ R\ (T_{2}-T_{1})=\alpha \ n\ R\ T_{1}\ \left(\left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}-1\right)~.} Simplifying, T 2 − T 1 = T 1 ( ( P 2 P 1 ) γ − 1 γ − 1 ) , T 2 T 1 − 1 = ( P 2 P 1 ) γ − 1 γ − 1 , T 2 = T 1 ( P 2 P 1 ) γ − 1 γ . {\displaystyle {\begin{aligned}T_{2}-T_{1}&=T_{1}\ \left(\left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}-1\right)\ ,\\{\frac {T_{2}}{T_{1}}}-1&=\left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}-1\ ,\\T_{2}&=T_{1}\ \left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}~.\end{aligned}}} === Derivation of discrete formula and work expression === The change in internal energy of a system, measured from state 1 to state 2, is equal to At the same time, the work done by the pressure–volume changes as a result from this process, is equal to Since we require the process to be adiabatic, the following equation needs to be true By the previous derivation, Rearranging (c4) gives P = P 1 ( V 1 V ) γ . {\displaystyle P=P_{1}\ \left({\frac {V_{1}}{V}}\right)^{\gamma }~.} Substituting this into (c2) gives W = ∫ V 1 V 2 P 1 ( V 1 V ) γ d V . {\displaystyle W=\int _{V_{1}}^{V_{2}}P_{1}\ \left({\frac {V_{1}}{V}}\right)^{\gamma }\ \mathrm {d} V~.} Integrating we obtain the expression for work, W = P 1 V 1 γ V 2 1 − γ − V 1 1 − γ 1 − γ = P 2 V 2 − P 1 V 1 1 − γ . {\displaystyle W=P_{1}\ V_{1}^{\gamma }{\frac {V_{2}^{1-\gamma }-V_{1}^{1-\gamma }}{1-\gamma }}={\frac {P_{2}\ V_{2}-P_{1}\ V_{1}}{1-\gamma }}~.} Substituting γ = ⁠α + 1/α⁠ in second term, W = − α P 1 V 1 γ ( V 2 1 − γ − V 1 1 − γ ) . {\displaystyle W=-\alpha \ P_{1}V_{1}^{\gamma }\ \left(V_{2}^{1-\gamma }-V_{1}^{1-\gamma }\right)~.} Rearranging, W = − α P 1 V 1 ( ( V 2 V 1 ) 1 − γ − 1 ) . {\displaystyle W=-\alpha \ P_{1}\ V_{1}\ \left(\left({\frac {V_{2}}{V_{1}}}\right)^{1-\gamma }-1\right)~.} Using the ideal gas law and assuming a constant molar quantity (as often happens in practical cases), W = − α n R T 1 ( ( V 2 V 1 ) 1 − γ − 1 ) . {\displaystyle W=-\alpha \ n\ R\ T_{1}\ \left(\left({\frac {V_{2}}{V_{1}}}\right)^{1-\gamma }-1\right)~.} By the continuous formula, P 2 P 1 = ( V 2 V 1 ) − γ , {\displaystyle {\frac {P_{2}}{P_{1}}}=\left({\frac {V_{2}}{V_{1}}}\right)^{-\gamma }\ ,} or ( P 2 P 1 ) − 1 γ = V 2 V 1 . {\displaystyle \left({\frac {P_{2}}{P_{1}}}\right)^{-{\frac {1}{\gamma }}}={\frac {V_{2}}{V_{1}}}~.} Substituting into the previous expression for W, W = − α n R T 1 ( ( P 2 P 1 ) γ − 1 γ − 1 ) . {\displaystyle W=-\alpha \ n\ R\ T_{1}\ \left(\left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}-1\right)~.} Substituting this expression and (c1) in (c3) gives α n R ( T 2 − T 1 ) = α n R T 1 ( ( P 2 P 1 ) γ − 1 γ − 1 ) . {\displaystyle \alpha \ n\ R\ (T_{2}-T_{1})=\alpha \ n\ R\ T_{1}\ \left(\left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}-1\right)~.} Simplifying, T 2 − T 1 = T 1 ( ( P 2 P 1 ) γ − 1 γ − 1 ) , T 2 T 1 − 1 = ( P 2 P 1 ) γ − 1 γ − 1 , T 2 = T 1 ( P 2 P 1 ) γ − 1 γ . {\displaystyle {\begin{aligned}T_{2}-T_{1}&=T_{1}\ \left(\left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}-1\right)\ ,\\{\frac {T_{2}}{T_{1}}}-1&=\left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}-1\ ,\\T_{2}&=T_{1}\ \left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}~.\end{aligned}}} == Graphing adiabats ==