An ideal gas (γ = 1.40) expands slowly and adiabatically. If the final temperature is one third the initial temperature, by what factor does the...

The relationship between temperature and volume of an ideal gas, during adiabatic expansion or contraction, is defined by the following equation:

`TV^(gamma-1) = constant`

where, T and V are the temperature and volume of the ideal gas. This means that

`T_1V_1^(gamma-1) = T_2V_2^(gamma-1)`

or, `T_1/T_2 = (V_2/V_1)^(gamma-1)`

where, subscripts 1 and 2 denote the initial and final states. Here, the final temperature is 1/3 of initial temperature, that is T2 = 1/3 T1 or, T1 =...

The relationship between temperature and volume of an ideal gas, during adiabatic expansion or contraction, is defined by the following equation:

`TV^(gamma-1) = constant`

where, T and V are the temperature and volume of the ideal gas. This means that

`T_1V_1^(gamma-1) = T_2V_2^(gamma-1)`

or, `T_1/T_2 = (V_2/V_1)^(gamma-1)`

where, subscripts 1 and 2 denote the initial and final states. Here, the final temperature is 1/3 of initial temperature, that is T2 = 1/3 T1 or, T1 = 3 T2

Substituting this relation into the previous equation, we get:

`(V_2/V_1)^(gamma-1) =` (3 T2/T2) = 3

substituting the value of `gamma` in the equation, we get: 

`(V_2/V_1) = 3^(1/(gamma-1)) = 3^(1/(1.4-1)) = 3^(1/0.4) = 15.6`

Thus, the final volume of the gas is 15.6 times the initial gas volume. In other words, the volume changes by a factor of 15.6.

Hope this helps.