Welcome to our community

Be a part of something great, join today!

Maths in Temperature and ideal gas Thermometer

Dhamnekar Winod

Active member
Nov 17, 2018
149
Hi,
I didn't understand the maths involved in the below article in regard to temperature and ideal gas thermometer. If any member knows it, may reply me.




1602763019755.png

1602763122516.png
If triple point of water is fixed at 273.16 K, and experiments show that freezing point of air-saturated water is 273.15 K at 1 atm system pressure.(so, melting point of ice is also 273.15 K , then how triple point of water is $0.10^\circ C$ What is meant by real gas volume at thermal equilibrium with a system whose true temperature is $V_T$ be V. In this Math symbol$\big(\frac{\partial V}{\partial T}\big)_P$ what does subscript P indicate? My guess P means Partial.

I am stuck here. If get the answers to my questions satisfactorily, i shall proceed further.
 

Attachments

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,207
What, specifically, is it that you don't understand?

-Dan
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,416
If triple point of water is fixed at 273.16 K, and experiments show that freezing point of air-saturated water is 273.15 K at 1 atm system pressure.(so, melting point of ice is also 273.15 K , then how triple point of water is $0.10^\circ C$
$0\,^\circ C$ is defined to be the freezing point of water at the standard 1 atm pressure.
It corresponds to $273.15\,K$.
Since the triple point of water is at $273.16\,K$, then that means that it corresponds to $0.01\,^\circ C$.

What is meant by real gas volume at thermal equilibrium with a system whose true temperature is $V_T$ be V. In this Math symbol$\big(\frac{\partial V}{\partial T}\big)_P$ what does subscript P indicate? My guess P means Partial.

I am stuck here. If get the answers to my questions satisfactorily, i shall proceed further.
The subscript $P$ in the expression $\big(\frac{\partial V}{\partial T}\big)_P$ stands for the pressure $P$.
It means that we take the derivative of $V$ with respect to $T$ while we keep the pressure $P$ constant.

We have $V=\frac{RT}{P}$ for an ideal gas.
When we take the derivative, we treat $P$ as a constant so that $\big(\frac{\partial V}{\partial T}\big)_P=\big(\frac{\partial}{\partial T}\frac{RT}{P}\big)_P=\frac{R}{P}$.
 
Last edited: