When pressure changes from 1 atm to 2 atm, the change in pressure
`P = 2 atm`
`= 2 xx 1 xx 10^(5) N//m^(2)`
The force exerted on the piston
`F = PA = 2 xx 10^(5) xx 4 xx 10^(-4)`
`= 80 N`
The compression of the spring
`x = (F)/(k) = (80)/(10^(4)) = 0.008 m`
The change in volume of air due to displacement of piston by `x`
`Delta V = Ax = 4 xx 10^(-4) xx 0.008`
`= 3.2 xx 10^(-6) m^(3)`
`:.` Final volume, `V_(2) = V_(1) + Delta V`
`= 20 xx 10^(-6) + 3.2 xx 10^(-6)`
By equation of state
`(P_(1) V_(1))/(T_(1)) = (P_(2) V_(2))/(T_(2))`
`T_(2) = (P_(2) V_(2) T_(1))/(P_(1) V_(1)) = ((3)/(1)) xx ((23.3 xx 10^(6)))/((20 xx 10^(-6))) xx (273 + 20)`
`= 1020 K`
The change in internal energy air
`Delta U = mC_(v) Delta T`
`= (2.38 xx 10^(-5)) xx 718 xx (1020 - 293)`
`= 12.42 J`
Work done in compressing the spring by `x`
`W = (1)/(2) kx^(2) = (10^(4))/(2) xx (0.008)^(2) = 0.32 J`
From the first law of thermodynamics
`Q = Delta U + W = 12.42 + 0.32 = 1274 J`