By integration, make expressions of work done by gas in
(a) Isobaric process (p=constant)
(b) Isothermal process (pV=constant)
(c) Adiabatic process (`PV^gamma=` constant)

Answer

A, D

Explanation

(a) Isobaric process
`W=int_(V_i)^(V_f)pdV=p int_(V_i)^(V_f)dV` (as p= constant)
`=p[V]_(V_i)^(V_f)=p(V_f-V_i)`
`=pDeltaV`
(b) Isothermal process
`W=int_(V_i)^(V_f)pdV=int_(V_i)^(V_f)((nRT)/(V))dV` (as `p=(nRT)/(V)`)
`=nRTint_(V_i)^(V_f)(dV)/(V)` (as T=constant)
`=nRTIn((V_f)/(V_i))=nRTIn((p_i)/(p_f))` (as `p_iV_i=p_fV_f` So, `V_f/V_i=p_i/p_f`)
(c) Adiabatic process
`pV^gamma=` constant=k(say)=`p_iV_i^gamma=p_fV_f^gamma`
Further, `p=(k)/(V^gamma)=kV^-gamma`
`W=int_(V_i)^(V_f)pdV=int_(V_i)^(V_f)kV^(-gamma)dV=[(kV^(gamma+1))/(-gamma+1)]_(V_i)^(V_f)`
`=(kV_f^(-gamma+1)-kV_i^(-gamma+1))/(-gamma+1)=(p_fV_f^gammaV_f^(-gamma+1)-p_iV_i^gammaV_i^(-gamma+1))/(1-gamma)`
`=(p_fV_f-p_iV_i)/(1-gamma)=(nRT_f-nRT_i)/(1-gamma)=(nRDeltaT)/(1-gamma)`