## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Tip our Team

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

## Holooly Tables

All the data tables that you may search for.

## Holooly Help Desk

Need Help? We got you covered.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Help Desk

Need Help? We got you covered.

## Q. 2.10

Determine whether the following quantities are thermodynamic properties or not :

(a)$\left(\frac{v}{T}\right) d p+\left(\frac{P}{T}\right) d v$ (b)$\left(\frac{d T}{T}\right)+\left(\frac{p}{T}\right) d v$

(c) $\left(\frac{d T}{T}\right)+\left(\frac{p}{v}\right) d v$ (d) $\left(\frac{d T}{T}\right)-\left(\frac{v}{T}\right) d p$

## Verified Solution

A quantity dz = Mdx + Ndy would be an exact differential, and hence a thermodynamic property, if

$\left(\frac{\partial M}{\partial y}\right)_x=\left(\frac{\partial N}{\partial x}\right)_y$

(a) Here $M=\frac{v}{T}=\frac{R}{p}$

(pv = RT)

\begin{aligned}&M=\frac{v}{T}=\frac{R}{p} \\&N=\frac{p}{T}=\frac{R}{v}\end{aligned}

x = p and y = v

\begin{aligned}\left(\frac{\partial M}{\partial y}\right)_x &=\left[\left(\frac{\partial}{\partial v}\right)\left(\frac{R}{p}\right)\right]_p=0 \\\left(\frac{\partial N}{\partial x}\right)_y &=\left[\left(\frac{\partial}{\partial p}\right)\left(\frac{R}{v}\right)\right]_v=0\end{aligned}

Since $\left(\frac{\partial M}{\partial y}\right)_x=\left(\frac{\partial N}{\partial x}\right)_y$,the given expression is an exact differential, and hence thermodynamic property.

(b) Here  $M=\frac{1}{T}$

$N=\frac{p}{T}$

x = T and y = v

\begin{aligned}\left(\frac{\partial M}{\partial y}\right)_x &=\left[\left(\frac{\partial}{\partial v}\right)\left(\frac{1}{T}\right)\right]_T=0 \\\left(\frac{\partial N}{\partial x}\right)_y &=\left[\left(\frac{\partial}{\partial T}\right)\left(\frac{R}{v}\right)\right]_v=0\end{aligned}

$\left(\frac{p}{T}=\frac{R}{v}\right)$

Since $\left(\frac{\partial M}{\partial y}\right)_x=\left(\frac{\partial N}{\partial x}\right)_y$, the given expression is an exact differential, and hence a thermodynamic property.

(c) Here $M=\frac{1}{T}$

$N=\frac{p}{v}$

x = T and y = v

\begin{aligned}&\left(\frac{\partial M}{\partial y}\right)_x=\left[\left(\frac{\partial}{\partial v}\right)\left(\frac{1}{T}\right)\right]_T=0 \\&\left(\frac{\partial N}{\partial x}\right)_y=\left[\left(\frac{\partial}{\partial T}\right)\left(\frac{p}{v}\right)\right]_v \neq 0\end{aligned}

Since $\left(\frac{\partial M}{\partial y}\right)_x \neq\left(\frac{\partial N}{\partial x}\right)_y$, the given expression is not an exact differential, and hence not a thermodynamic property

(d) Here $M=\frac{1}{T}$

$N=\frac{v}{T}$

x = T and y = p

\begin{aligned}\left(\frac{\partial M}{\partial y}\right)_x &=\left[\left(\frac{\partial}{\partial p}\right)\left(\frac{1}{T}\right)\right]_T=0 \\\left(\frac{\partial N}{\partial x}\right)_y &=\left[\left(\frac{\partial}{\partial T}\right)\left(\frac{R}{P}\right)\right] p=0\end{aligned}

$\left(\because \frac{v}{T}=\frac{R}{p}\right)$

Since $\left(\frac{\partial M}{\partial y}\right)_x=\left(\frac{\partial N}{\partial x}\right)_y$, the given expression is an exact differential, and hence a thermodynamic property.