In an adiabatic process, the system is insulated from the surroundings and heat absorbed or released is zero. From Eq. (12.1), we see that work done by the gas results in decrease in its internal energy (and hence its temperature for an ideal gas).
We quote without proof (the result that you will learn in higher courses) that for an adiabatic process of an ideal gas.
`color{purple} { P V^ γ = "const"}` ......................... (12.13)
where γ is the ratio of specific heats (ordinary or molar) at constant pressure and at constant volume.
where `γ` is the ratio of specific heats (ordinary or molar) at constant pressure and at constant volume.
`color{green} {gamma = (C_p)/C_v}`
Thus if an ideal gas undergoes a change in its state adiabatically from `(P_1, V_1)` to `(P_2, V_2)` :
`color{purple} {P_1 V_1^γ = P_2 V_2^ γ}` ........................ (12.14)
Figure12.8 shows the P-V curves of an ideal gas for two adiabatic processes connecting two isotherms.
We can calculate, as before, the work done in an adiabatic change of an ideal gas from the state `color{green} {(P_1, V_1, T_1)}` to the state `color{purple} { (P_2, V_2, T_2)}`.
`color{navy} { W = int_(V_1) ^(V_2) P dV`
`= "constant" xx int_(V_1) ^(V_2) (dV)/V^(gamma) = "constant" xx (V^(- gamma + 1))/(1 - gamma)|_(V_1)^(V_2)`
`= ("constant")/(1 -gamma) xx { 1/(V_2^(gamma -1)) - 1/ (V_1^(gamma -1)) ]` ....................12.15
From Eq. (12.34), the constant is `P_1V_1^γ` or `P_2V_2^γ`
`W = 1/(1 -gamma) [ ( P_2V_2^gamma)/(V_2^(gamma -1) - P_1V_1^gamma)/ ( V_1^(gamma -1 ) ) ]`
`color{purple} {= 1/ (1- gamma ) [ ( P_2V_2 - P_1V_1 ] = (mu R(T_1 - T_2 ))/(gamma -1) ` ......................12.16
As expected, if work is done by the gas in an adiabatic process (W > 0), from Eq. (12.16), `T_2 < T_1`. On the other hand, if work is done on the gas (W < 0), we get `T_2 > T_1` i.e., the temperature of the gas rises.
In an adiabatic process, the system is insulated from the surroundings and heat absorbed or released is zero. From Eq. (12.1), we see that work done by the gas results in decrease in its internal energy (and hence its temperature for an ideal gas).
We quote without proof (the result that you will learn in higher courses) that for an adiabatic process of an ideal gas.
`color{purple} { P V^ γ = "const"}` ......................... (12.13)
where γ is the ratio of specific heats (ordinary or molar) at constant pressure and at constant volume.
where `γ` is the ratio of specific heats (ordinary or molar) at constant pressure and at constant volume.
`color{green} {gamma = (C_p)/C_v}`
Thus if an ideal gas undergoes a change in its state adiabatically from `(P_1, V_1)` to `(P_2, V_2)` :
`color{purple} {P_1 V_1^γ = P_2 V_2^ γ}` ........................ (12.14)
Figure12.8 shows the P-V curves of an ideal gas for two adiabatic processes connecting two isotherms.
We can calculate, as before, the work done in an adiabatic change of an ideal gas from the state `color{green} {(P_1, V_1, T_1)}` to the state `color{purple} { (P_2, V_2, T_2)}`.
`color{navy} { W = int_(V_1) ^(V_2) P dV`
`= "constant" xx int_(V_1) ^(V_2) (dV)/V^(gamma) = "constant" xx (V^(- gamma + 1))/(1 - gamma)|_(V_1)^(V_2)`
`= ("constant")/(1 -gamma) xx { 1/(V_2^(gamma -1)) - 1/ (V_1^(gamma -1)) ]` ....................12.15
From Eq. (12.34), the constant is `P_1V_1^γ` or `P_2V_2^γ`
`W = 1/(1 -gamma) [ ( P_2V_2^gamma)/(V_2^(gamma -1) - P_1V_1^gamma)/ ( V_1^(gamma -1 ) ) ]`
`color{purple} {= 1/ (1- gamma ) [ ( P_2V_2 - P_1V_1 ] = (mu R(T_1 - T_2 ))/(gamma -1) ` ......................12.16
As expected, if work is done by the gas in an adiabatic process (W > 0), from Eq. (12.16), `T_2 < T_1`. On the other hand, if work is done on the gas (W < 0), we get `T_2 > T_1` i.e., the temperature of the gas rises.