Question

A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure `P_i=10^5` Pa and volume `V_i=10^-3m^3` changes to a final state at `P_f=(1//32)xx10^5Pa and V_f=8xx10^-3m^3` in an adiabatic quasi-static process, such that `P^3V^3=constant.` Consider another thermodynamic process that brings the system form the same initial state to the same final state in two steps: an isobaric expansion at `P_i` followed by an isochoric (isovolumetric ) process at volume `V_r.` The amount of heat supplied to the system i the two-step process is approximately
A. `112J`
B. `294J`
C. `588J`
D. `813J`

Answer 1

Correct Answer - C
(c)`P^3V^5=constant rArr PV^(5//3)=constant rArr gamma=5/3`
`rArr` monoatomic gas
For adiabatic process

`W=(P_fV_f-P_iV_i)/(1-gamma)=(1/32xx10^5xx8xx10^-3-10^5xx10^-3)/(1-5/3)`
`:. W=(25-100)/((3-5)//3)=(75xx3)/2=112.5J`
From first low of thermodynamics `q=DeltaU+w :. DeltaU=-w`
`:. DeltaU=-112.5J`
Now applying first law of thermodynamics for process
1&2 and adding `q_1+q_2=DeltaU+P_i(V_f-V_i)`
`=-112.5+10^5(8-1)xx10^-3=587.55`