A masslesss piston divides a closed thermallyy insulated cylinder into two equal parts. One part contains M = 28 g of nitrogen. At this temperature, one-third of molecules are dissociated into atoms and the other part is evacuated. The piston is released and the gas fills the whole volume of the cylinder at temperature `T_(0)`. Then, the piston is slowly displaced back to its initial position. calculate the increases in internal energy of the gas. Neglect further dissociation of molecules during, the motion of the piston.

Explanation

Molar mass of `N_(2)` gas is M = 28 g
Number of moles of `N_(2)`, `n_(0) = (m//M) = 1`
One-third molecules are dissociated into atoms.
No. of moles of diatomic `N_(2) // 2//3`
No. of moles of monatomic nitrogen `= 2 xx (1//3)`
For monatomic gas `C_(v_(1)) = (3//2) R`
For diatomic gas `C_(v_(2)) = (5//2) R`
Average molar specific heat at constant volume `C_(v)` is
`(n_(1) C_(v_(1)) + n_(2) C_(v_(2))) /(n_(1)p0 + n_(2)) = (((2)/(2)) ((3)/(2)) R + ((2)/(3)) ((5)/(2))R)/((2)/(3) + (2)/(3)) = 2 R`
`C_(P) - C_(v) = R implies C_(P) = 3 R`
`gamma = (C_(P))/(C_(V)) = 1.5`
Piston is diaplaced bac to its initial position, and during the adiabatic compression, volume of the gas mixture is halved.
For adiabatic compression,
`T_(1) V_(1)^(gamma -1) = T_(2) V_(2)^(gamma -1)`
Where `T_(1) = T_(0) , V_(1) = V_(0)`
`T_(2) = ? , V_(2) = V_(0)//2`
`T_(0) V_(0)^(gamma - 1) = T_(2) ((V_(0))/(2))^(gamma -1) implies T_(2) = T_(0) sqrt2`
`Delta U = (n_(1) + n_(2)) C_(V) (T_(2) - T_(0))`
`= (4)/(3) xx 0.2 R (T_(0) sqrt2 - T_(0))`
`Delta U = (8)/(3) RT_(0) (sqrt2 - 1)`