By Sunil Bhardwaj

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In an Adiabatic process heat neither enters nor leaves the system i.e. $$q = 0$$ From 1st law of thermodynamics at constant pressure, $$q = \Delta E + P\Delta V$$ $$\therefore 0 = \Delta E + P\Delta V$$ or $$\Delta E = -P\Delta V\qquad ....(1)$$ for an infinitesimal small change in a reversible process we can write, $$dE = -PdV \qquad ....(2)$$ For one mole of ideal gas $$PV = RT \qquad \therefore P = \frac { RT }{ V }$$ and $${ C }_{ v } = \frac { dE }{ dT } \qquad \therefore dE = { C }_{ v }dT$$ Lets substitute these two values in equation (2) $$\therefore { C }_{ v }dT = -\frac { RT }{ V } dV$$ on rearranging this equation, $${ C }_{ v }\frac { dT }{ T } = -R\frac { dV }{ V }$$ This equation can be integrated between the limits from $${ T }_{ 1 }$$ to $${ T }_{ 2 }$$ and $$V_{ 1 }$$ to $$V_{ 2 }$$. $${ C }_{ v } \int _{ { T }_{ 1 } }^{ { T }_{ 2 } }{ \frac { dT }{ T } } = -R \int _{ V_{ 1 } }^{ { V }_{ 2 } }{ \frac { dV }{ V } }$$ $$\therefore { C }_{ v } ln\frac { { T }_{ 2 } }{ { T }_{ 1 } } = -R ln\frac { V_{ 2 } }{ V_{ 1 } } = R ln\frac { V_{ 1 } }{ V_{ 2 } }$$ $$\therefore { C }_{ v } ln\frac { { T }_{ 2 } }{ { T }_{ 1 } } = \left( { C }_{ p } - { C }_{ v } \right) ln\frac { V_{ 1 } }{ V_{ 2 } } \qquad ....\left( R = { C }_{ p } - { C }_{ v } \right)$$ or $$ln\frac { { T }_{ 2 } }{ { T }_{ 1 } } = \left( \frac { { C }_{ p } - { C }_{ v } }{ { C }_{ v } } \right) ln\frac { V_{ 1 } }{ V_{ 2 } }$$ $$= \left( \frac { { C }_{ p } }{ { C }_{ v } } - 1 \right) ln\frac { V_{ 1 } }{ V_{ 2 } }$$ $$= \left( \gamma - 1 \right) ln\frac { V_{ 1 } }{ V_{ 2 } } \qquad ....where \ \gamma = \frac { { C }_{ p } }{ { C }_{ v } }$$ $$\therefore \frac { { T }_{ 2 } }{ { T }_{ 1 } } = { \left( \frac { V_{ 1 } }{ V_{ 2 } } \right) }^{ \left( \gamma - 1 \right) } \qquad ....(3)$$ $$or \ { T }_{ 1 }{ V }_{ 1 }^{ \left( \gamma - 1 \right) } = { T }_{ 2 }{ V }_{ 2 }^{ \left( \gamma - 1 \right) }$$ In general, The relation between temperature and volume is, $$\boxed { { T }{ V }^{ \left( \gamma - 1 \right) } = { T }{ V }^{ \left( \frac { R }{ { C }_{ v } } \right) } = Constant } \qquad ....(4)$$ Also for two states of one mole of an ideal gas, $${ P }_{ 1 }V_{ 1 } = R{ T }_{ 1 }$$ and $${ P }_{ 2 }V_{ 2 } = R{ T }_{ 2 }$$ $$\therefore \boxed { \frac { { T }_{ 2 } }{ { T }_{ 1 } } = { \frac { { P }_{ 2 } }{ { P }_{ 1 } } }{ \frac { V_{ 2 } }{ V_{ 1 } } } }$$ and $$\boxed { \frac { { V }_{ 1 } }{ V_{ 2 } } = { \frac { { P }_{ 2 } }{ { P }_{ 1 } } \frac { T_{ 1 } }{ T_{ 2 } } } }$$ lets substitute these in equation (3) $$\boxed { \therefore \frac { { P }_{ 2 }V_{ 2 } }{ { P }_{ 1 }V_{ 1 } } = { \left( \frac { V_{ 1 } }{ V_{ 2 } } \right) }^{ \left( \gamma - 1 \right) }}$$ $$\therefore { P }_{ 2 }V_{ 2 }{ \left( V_{ 2 } \right) }^{ \left( \gamma - 1 \right) } = { P }_{ 1 }V_{ 1 }{ \left( V_{ 1 } \right) }^{ \left( \gamma - 1 \right) }$$ or $${ P }_{ 2 }V_{ 2 }^{ \gamma } = { P }_{ 1 }V_{ 1 }^{ \gamma }$$ In general, The relation between pressure and volume is, $$\boxed { { P }V^{ \gamma } = Constant }$$ $$\boxed { \frac { { V }_{ 1 } }{ V_{ 2 } } = { \frac { { P }_{ 2 }T_{ 1 } }{ { P }_{ 1 }T_{ 2 } } }}$$ $$\therefore \frac { { T }_{ 2 } }{ { T }_{ 1 } } = { \left( { \frac { { P }_{ 2 }T_{ 1 } }{ { P }_{ 1 }T_{ 2 } } } \right) }^{ \left( \gamma - 1 \right) }$$ $$\therefore \frac { { T }_{ 2 }^{ \gamma } }{ { T }_{ 1 }^{ \gamma } } = { \left( { \frac { { P }_{ 2 } }{ { P }_{ 1 } } } \right) }^{ \left( \gamma - 1 \right) }$$ or $$\boxed { \frac { { T }_{ 2 } }{ { T }_{ 1 } } = { \left( { \frac { { P }_{ 2 } }{ { P }_{ 1 } } } \right) }^{ \left( \frac { \gamma - 1 }{ \gamma } \right) } }$$ This is the relation between temperature and pressure.

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