Question

Given $d\varphi =f\left(T\right)dT+\left(T/V\right)dVandd\psi =Tdp+\left(T/p^2\right)dV,$ then

• A. both $\varphi$ and $\psi$ are properties
• B. neither $\varphi$ nor $\psi$ is a property
• C. $\varphi$ is a property but $\psi$ is not a property
• D. $\psi$ is a property but $\varphi$ is not a property

Answer 1

The correct option is B neither $\varphi$ nor $\psi$ is a property

$d\psi =Tdp+\frac{T}{p^2}dV$

$\psi =f(p,V)$

$d\psi ={\left(\frac{\partial \psi }{\partial p}\right)}_{V}dp+{\left(\frac{\partial \psi }{\partial V}\right)}_{p}dV$

${\left(\frac{\partial \psi }{\partial p}\right)}_V=M=T;{\left(\frac{\partial \psi }{\partial V}\right)}_p=N=\frac{T}{p^2}$

$\left(\frac{\partial M}{\partial V}\right)=0;\left(\frac{\partial N}{\partial p}\right)=\frac{-2T}{p^3}$

$\frac{\partial M}{\partial V}\ne \frac{\partial N}{\partial p}\to hence\psi isnotaproperty$

$d\varphi =f\left(T\right)dt+\frac{T}{V}dV$

$\varphi =f(T;V)$

$d\varphi ={\left(\frac{\partial \varphi }{\partial T}\right)}_{V}dT+{\left(\frac{\partial \varphi }{\partial V}\right)}_{T}dV$

${\left(\frac{\partial \varphi }{\partial T}\right)}_V=f\left(T\right)=M$

$\left(\frac{\partial M}{\partial V}\right)=0$

${\left(\frac{\partial \varphi }{\partial V}\right)}_T=\frac{T}{V}=N$

$\frac{\partial N}{\partial T}=\frac{1}{V}$

$\frac{\partial M}{\partial V}\ne \frac{\partial N}{\partial T}\to Hence\varphi isalsonotaproperty$