Figure. Shows three processes for an ideal gas. The temperature at 'a' is 600 K, pressure of 1 atm, and volume 1 L. The volume at 'b' is 4 L. Out of two processes, ab and ac, one is adiabatic and other is isothermal. The ratio of specific heats of the gas is 1.5.
a. Which of the processes ab and ac is adiabatic ? Why
b. Compute the pressure of the gas at b and c.
c. Compute the temperature at b and c.
d. Complete the volume at c.

Explanation

a. The process ab is adiabatic because it is steeper than ac.
b. `P_(a) = 16 "atm", V_(b) = 4L , V_(a) = 1 L`
`implies T_(a) = 600 K`
Process 'ab' (adiabatic)
`implies P_(a) V_(a)^(gamma) = P_(b) V_(b)^(gamma)`
`P_(b) = P_(a) ((V_(a))/(V_(b)))^(gamma) = 16 xx ((1)/(4))^(1.5)`
`implies P_(b) = 2 "atm"`
As bc is isobaric,
`P_(b) = P_(c ) = 2 "atm"`
c. `T_(c ) = T_(a) = 600 K` because ac is isothermal
Process 'ab' is adiabatic hence
`T_(b)` can be calculated using `T_(a) V_(a)^(gamma - 1) = T_(b) T_(b)^(gamma -1)`
`implies ((P_(a) V_(a))/(T_(a))) = ((P_(b) V_(b))/(T_(b)))`
Hence, `T_(b) = T_(a) ((V_(a))/(V_(b)))^(gamma - 1) = 600 ((1)/(4))^(0.5) = 300 K`
d. Process 'ac' (isothermal)
`implies P_(a) V_(a) = P_(c ) V_(c )`
`V_(c ) = (P_(a) V_(a))/(P_(c)) = ((16 xx 1))/(2) = 8 L`