Question

One mole ideal monoatomic gas is heated according to path AB and AC. If temperature of state B and C are equal. Calculate $\frac{q_{AC}}{q_{AB}}\times 10$

Answer 1

From the graph for the proccess AC, we get that $P=KV\left(Kisaproportionalityconstant\right)P{V}^{-1}=K.......\left(1\right)$
This equation satisfies the equation of polytropic proccess, i.e. $P{V}^n=Constant.........\left(2\right)$
From the equation (1) and (2) we get $n=-1$
So the process AC is a polytropic process.
For the polytropic proccess molar heat capacity is $C_m=C_{v,m}+\frac{R}{1-n}\Rightarrow C_m=C_v+\frac{R}{2}=2R(n=-1)$
Process AB is isobaric $C_m=C_p=\frac{5R}{2}$
$\frac{q_{AC}}{q_{AB}}=\frac{\underset{T_A}{\overset{T_C}{\int }}n{C}_{m}dT}{\underset{T_A}{\overset{T_B}{\int }}n{C}_{p,m}dT}=\frac{\underset{T_A}{\overset{T_C}{\int }}2RdT}{\underset{T_A}{\overset{T_B}{\int }}\frac{5R}{2}dT}=\frac{2R(T_C-T_A)}{2.5R(T_B-T_A)}=0.8(As{T}_C=T_B,then{T}_C-T_A=T_C-T_B)$