An ideal gas at initial temperature `T_(9)` and initial volume `V _(0)` is expanded adiabatically to a volume `2V_(0).` The gas is then ecp[anded isothermally to a volume `5V_(0)` and there after compressed adiabatically so that the temperature of tha gas becomes again `T_(0).` If the final volume of the gas is `alphaV_(0)` then the value of constant `alpha` is

Options

A. `2.5`
B. `1.5`
C. `2.0`
D. `3.0`

Answer

A

Explanation

Given,
For an ideal gas,
Initial temperature, `T_(1) =T_(0)`
Initiaol volume, `V_(1) = V_(0)`
After expanded adiabatically,
Volume of gas,` V_(2) = 2V_(0)`
Then after expanding isothermally,
Volume of gases, `V_(3)= 5V_(0)`
`T_(3) = T_(0)`
Final volume of gas ` = alpha V_(0)`
Since, gas is expanded adiabatically,
`T _(1) V_(1) ^(gamma -1) = T_(2) V_(2) ^(gamma -1)`
`T _(0) V_(0) ^( gamma -1) = T_(2) (2V_(0))^(gamma -1)`
`implies T_(2) = (T_(0))/(2 ^(gamma -1))`
The adiabatically expanded gas is again expanded isothermally
`i.e. P _(1) V_(2) = P_(2) V_(2)`
` = (P_(1))/(P _(2)) = (5V_(0))/(2 V_(0))`
`implies (P_(1))/(P_(2)) = 5/2`
This gas is compressed adiabaticaly, Then,
`implies T_(2) V_(3) ^(gamma -1) = T_(3) V_(4) ^(gamma -1) `
`implies (T_(0))/(2 ^(gamma -1)) (5V_(0)) ^(gamma -1) = T_(0) (V_(4)) ^( gamma -1)`
`((5 V_(0))/(2 ^(gamma -1))) ^(gamma -1) = V_(4) ^(gamma -1)`
`(5V_(0))/(2) = V_(4)`
` V_(3) = 2.5 V _(0)`
` therefore alpha = 2.5`