Consider that an ideal gas (n moles) is expanding in a process given by p = f (V), which passes through a point (V_0, p_0). Show that the gas is absorbing heat at (p_0, V_0) if the slope of the curve p =f (V) is larger than the slope of the adiabatic passing through (p_0,V_0).

Consider that an ideal gas (n moles) is expanding in a process given b

1.	Consider that an ideal gas (n moles) is expanding in a process given by p = f (V), which passes through a point (V_0, p_0). Show that the gas is absorbing heat at (p_0, V_0) if the slope of the curve p =f (V) is larger than the slope of the adiabatic passing through (p_0,V_0).
Answer» Solution :Slope of graph P=f(V) at POINT `(V_0, p_0)=f(V_0)` For adiabatic process `PV^(gamma)`=K (Constant) `therefore` At point `(V_0,P_0)P_0 V_0^gamma=k` `therefore P_0=k/V_0^gamma` `therefore dP_0=k(-gamma)V_0^(-gamma-1). dV_0` `therefore (dP_0)/(dV_0)=-gamma k/V_0^gamma xx 1/V_0` `therefore` Slope =`-gamma P_0/V_0`....(1)`[because k/V_0^gamma =P_0]` Now dQ=dU+dW `=nC_V dT + PdV`...(2) and PV =nRT `therefore T=(PV)/(nR)=(f(V)V)/(nR)` `therefore dT=1/(nR)[f(V)dV+Vf.(V)dV]` `=1/(nR) [f(V)+Vf.(V)]dV`...(3) `rArr` Equ. (2) `dQ=nC_V xx1/(nR)[f(V)+Vf.(V)]dV` `therefore ["dQ"/"dV"]_(V=V_0) =C_V/R[f(V_0)+V_0f.(V_0)]+f(V_0)` ...(4) Now, `C_P-C_V=R` `therefore C_P/C_V-1=R/C_V` `gamma-1=R/C_V` `therefore C_V/R=1/(gamma-1)`...(5) Putting value of (5) in (4), `((dQ)/(dV))_(V=V_0)=(C_V/R+1)f(V_0)+V_0f.(V_0)` `=[1/(gamma-1)+1]f(V_0) +(V_0f.(V_0))/(gamma-1)` `=gamma/(gamma-1)P_0+V_0/(gamma-1)f.(V_0) [ because f(V_0)=P_0]` When gas expended then `(dQ)/(dV) GT 0` and heat is absorbed. Now `(dQ)/(dV) gt 0` hence , `(gammaP_0)/(gamma-1)+V_0/(gamma-1) f.(V_0) gt0` `therefore gammaP_0+V_0f.(V_0) gt 0` `therefore f.(V_0)=-(gammaP_0)/V_0`