Consider that an ideal gas 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 adiabat passing through (P_0, V_0).

Answers (1)

Slope of graph at

(V_0,P_0)=\frac{dP}{dV}_{(P_0,V_0)}

P=f(V) for adiabatic process PV^{\gamma }=\text{constant }K

P=\frac{K}{V^{\gamma}}

\frac{dP}{dV}=-\frac{\gamma P}{}V

(\frac{dP}{dV})_{(P_0,V_0)}=-\gamma \frac{P_0}{V_0 }

P=f(V)

dQ=dU+dW

dQ=nC_vdT+PdV

PV=nRT

T=\frac{PV}{nR}=\frac{V}{nR}f(V)

\frac{dT}{dV}=\frac{1}{nR}[f(V)+Vf'(V)]

\frac{dQ}{dV}=nC_v\left (\frac{dT}{dV} \right )+P\left (\frac{dV}{dV} \right )=\frac{nC_v}{nR}[f(V)+V(f'(V))+P]

\frac{dQ}{dV}_{(V=V_0)}=\frac{C_v}{}R\left [f(V_0)+V_0(f'(V_0))+f(V_0) \right ]

=f(V_0)\left [\frac{C_v}{}R+1 \right ]+V_0f'(V_0)\frac{C_v}{}R

C_P-C_v=R\Rightarrow \frac{C_P}{C_V}-1=\frac{R}{C_v}

\gamma -1=\frac{R}{C_v}\Rightarrow C_v=\frac{R}{\gamma -1}\Rightarrow \frac{C_v}{}R=\frac{1}{\gamma -1}

\frac{dQ}{dV}_{(V=V_0)}=\frac{1}{r-1}\left [\gamma P_0+V_0f'{(V_0)} \right ]

\frac{dQ}{dV}_{(V=V_0)}>1 and \gamma >1 so\frac{1}{\gamma -1}>0

\gamma P_0+V_0f'(V_0)>0

V_0f'(V_0)>-\gamma P_0

f'V_0>-\frac{\gamma P_0}{V_0}

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