# Heat transfer and mercury

1. Mar 18, 2009

### Winzer

1. The problem statement, all variables and given/known data
The pressure on 0.01 litres of mercury is increased reversibly and isothermally
from zero to 1000 atm at room temperature (293 K). Mercury has a coefficient
of volume expansion β = 1.82 × 10−4 K−1 , and an isothermal compressibility
κT = 4.02 × 10−11 Pa−1 . Note: 1 atm= 1.013 × 105 Pa. Assuming that the
volume, V , changes very little, find
(i) how much heat is transferred in the compression;
(ii) the work done during the compression;
(iii) the change in internal energy.

2. Relevant equations
$$T dS= C_p dT -\beta V dP$$
$$\oint \frac{\def\dbar{{\mathchar'26\mkern-12mu d}Q} \dbar}{T}=0$$
$$\def\dbar{{\mathchar'26\mkern-12mu d}Q} \dbar=C_v dT$$

3. The attempt at a solution
I am attempting to find the final temp so I can implement: $$\def\dbar{{\mathchar'26\mkern-12mu d}Q} \dbar=C_v dT$$

Since the process is reversible, and under a complete cycle $$\oint \frac{\def\dbar{{\mathchar'26\mkern-12mu d}Q} \dbar}{T}=0$$ I set $$dS=0$$. Getting $$C_p \int \frac{dT}{T}=\beta T V \int dP$$ Is this correct so far?

2. Mar 18, 2009