An ideal monoatomic gas is confined in a cylinder by a spring-loaded piston of cross-section `8.0xx10^-3m^2`. Initially the gas is at 300K and occupies a volume of `2.4xx10^-3m^3` and the spring is in its relaxed (unstretched, unompressed) state, fig. The gas is heated by a small electric heater until the piston moves out slowly by 0.1m. Calculate the final temperature of the gas and the heat supplied (in joules) by the heater. The force constant of the spring is `8000 N//m`, atmospheric pressure is `1.0xx10^5 Nm^-2`. The cylinder and the piston are thermally insulated. The piston is massless and there is no friction between the piston and the cylinder. Neglect heat loss through lead wires of the heater. The heat capacity of the heater coil is negligible. Assume the spring to be massless.

Explanation

Initially, the pressure of the gas in the cylinder is atmosphere pressure as spring is in relaxed state. Therefore,
`P_(1)` atmospheric pressure `= 1.0 xx 10^(5) N//m^(2)`
`V_(1)`= initial volume `= 2.4 xx 10^(3) m^(3)`
`T_(1)` - initial temperature = 300 K
When the heat is supplied by the heater, the piston is compressed by 0.1 m. The reaction force of compression of spring is equal to kx which acts on the piston or on the gas as
`F = kx = 8000 xx 0.1 xx 800 N`
Pressure exerted on the piston by the spring is
`Delta P = (F)/(A) = (800)/(8 xx 10^(-3)) = 1 xx 10^(5) N//m^(2)`
The total pressure `P_(2)` of the gas inside cylinder is
`P_(2) = P_(atm) + Delta P =1 xx 10^(5) + 1 xx 10^(5) = 2 xx 10^(6) N//m^(2)`
Since the piston has moved outwards, there has been an increase of `Delta V` in the volume of the gas, i.e.,
`Delta V = A xx x = (8 xx 10^(-3)) xx (0.1)`
`= 8 xx 10^(-4) m^(3)`
The final volume of the gas
`V_(2) = V_(1) + Delta V = 2.4 xx 10^(-3) + 8 xx 10^(-4) = 3.2 xx 10^(-3) m^(3)`
Let `T_(2)` be the final temperature of gas. Then
`(P_(1) V_(1)//T_(1)) = (P_(2) V_(2)//T_(2)) implies T_(2) (P_(2) V_(2)//P_(1) V_(1)) T_(1)`
`= 300 xx (2 xx 10^(5) xx 3.2 xx 10^(-3))//(10^(5) xx 2.4 xx 10^(-3))`
`= 800 K`
Let the heat supplied by the heater be Q. This is used in two parts, i.e., a part is used in doing external work W due to expansion of the gas and the other part is used in increasing h internal energy of the gas. Hence,
`Q = W + Delta U`
Now `W = int_(V_(1))^(V_(2)) PdV = int_(0)^(x) (P_(atm) + (kx)/(A)) Adx`
[as pressure is `(P_(atm + kx//A` and `dV = Adx)]`
or `W = P_(atm) Ax + ((kx^(2))/(2))`
`= [10^(5) xx (8 xx 10^(-3)) (0.1) + (8000 xx (0.1)^(2))/(2)] = 120 J`
Further, `Delta = nC_(v) Delta T`
Number of moles of gas can be obtained from initial conditions and gas law is
`n = ((PV)/(RT)) = ((1 xx 10^(5) xx 2.4 xx 10)/(8.314 xx 300)) = 0.096 mol`
Thus, change in internal energy of gas is given as
`Delta U = n ((3)/(2) R) Delta T` (as for monatomic gas `C_(V) = (3)/(4) R`)
or `U = 0.906 xx (3)/(2) xx 8.314 xx 598.6 J`
Heat supplied by the heater = (120 + 598.6) = 718.6 J.