One mole of a perfect gas, initally at a pressure and temperature of `10^(5) N//m^(2)` and 300 K, respectively, expands isothermally unitl its volume is doubled and then adiabatically until its volume is again doubled. Find final pressure and temperature of the gas Find the total work done during the isothermal and adiabatic processes. Given `gamma = 1.4`. Also draw the P-V diagram for the process.

Explanation

Let `P_(1), V_(1), T_(1)` be the initial pressure, volume and the temperature of the gas, respectively. For isothermal expansion (1 to 2), we have
`V_(2) = 2V_(1) , P_(1) V_(1) = P_(2) V_(2)`
`P_(2) = P_(1) (V_(1)//V_(2)) = P_(1) // 2 = 0.5 xx 10^(5) N//m^(2)`
For adiabatic expension (2 to 3),
`P_(2) V_(2)^(gamma) = P_(3) V_(3)^(gamma)`
`implies V_(3) = 2V_(2)`
`implies P_(3) = P_(2) ((V_(2))/(V_(3)))^(gamma) = 0.5 xx 10^(5) xx (0.5)^(1.4)`
`implies P_(3) = 1.9 xx 10^(4) N//m^(2)`
`T_(2) V_(2)^(gamma -1) = T_(3) V_(3)^(gamma -1)`
`implies T_(3) = T_(3) ((V_(2))/(V_(3)))^(gamma - 1) = 300 ((1)/(4))^(0.4) = 227.35 K`
Hence, final pressure and temperatures are `1.9 xx 10^(4) N//m^(2)` and 227.35 K, respectively.
Work done is `W = W_(1 rarr2) + W_(2 rarr 3)`.