How Does Mercury's Compression Affect Heat Transfer and Energy?

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SUMMARY

The discussion focuses on the thermodynamic analysis of compressing 0.01 liters of mercury isothermally and reversibly at room temperature (293 K) under a pressure increase from 0 to 1000 atm. Key parameters include the coefficient of volume expansion (β = 1.82 × 10−4 K−1) and isothermal compressibility (κT = 4.02 × 10−11 Pa−1). The participants aim to calculate the heat transferred, work done, and change in internal energy during this process, emphasizing the application of the first law of thermodynamics and relevant equations.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically the first law of thermodynamics.
  • Familiarity with concepts of isothermal processes and reversible compression.
  • Knowledge of the coefficients of volume expansion and isothermal compressibility.
  • Ability to manipulate thermodynamic equations, including those involving heat transfer and work done.
NEXT STEPS
  • Study the first law of thermodynamics in detail, focusing on its application to isothermal processes.
  • Learn about the derivation and implications of the coefficients of volume expansion and isothermal compressibility.
  • Explore the relationship between heat transfer and work done in thermodynamic cycles.
  • Investigate the implications of compressibility in real gases versus ideal gases.
USEFUL FOR

This discussion is beneficial for students and professionals in thermodynamics, particularly those studying heat transfer, energy systems, and fluid mechanics. It is especially relevant for physicists and engineers working with compressible fluids like mercury.

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Homework Statement


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.

Homework Equations


T dS= C_p dT -\beta V dP
\oint \frac{\def\dbar{{\mathchar&#039;26\mkern-12mu d}Q} <br /> \dbar}{T}=0
\def\dbar{{\mathchar&#039;26\mkern-12mu d}Q} <br /> \dbar=C_v dT

The Attempt at a Solution


I am attempting to find the final temp so I can implement: \def\dbar{{\mathchar&#039;26\mkern-12mu d}Q} <br /> \dbar=C_v dT

Since the process is reversible, and under a complete cycle \oint \frac{\def\dbar{{\mathchar&#039;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?
 
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Your first and last equations doesn't look right. Check your units and your T's.
 
I'd start with the first law.
 

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