Find Work done by block of gold?

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SUMMARY

The discussion focuses on calculating the work done by a block of gold when heated from 20°C to 1064°C at a pressure of 1.0 atm. The density of solid gold is 19.3 g/cm³, while the density of liquid gold at 1064°C is 17.3 g/cm³. The user initially calculated the volume using an incorrect density difference, leading to an erroneous work calculation of 20619 J. The correct approach involves determining the volumes at both temperatures and applying the work formula W = -pΔV accurately.

PREREQUISITES
  • Understanding of density calculations (density = mass/volume)
  • Knowledge of the ideal gas law (PV = nRT)
  • Familiarity with work done in thermodynamics (W = -pΔV)
  • Basic conversion skills between units (e.g., g/cm³ to kg/m³)
NEXT STEPS
  • Calculate the volume of solid gold at 20°C using its density of 19.3 g/cm³.
  • Calculate the volume of liquid gold at 1064°C using its density of 17.3 g/cm³.
  • Re-evaluate the work done using the correct volume change (ΔV) in the formula W = -pΔV.
  • Explore the implications of phase changes on density and work calculations in thermodynamic processes.
USEFUL FOR

This discussion is beneficial for students studying thermodynamics, particularly those focusing on phase changes and work calculations in physical chemistry or engineering contexts.

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Homework Statement


The density of solid gold at 20°C is
ρ = 19.3 g/cm3.
When it is liquid at 1064°C the density of the liquid state is decreased: 17.3 g/cm3. How much work does a block of gold of mass 407 kg do if it is heated at
p = 1.0 atm
from 20°C to 1064°C?

Homework Equations


density=mass/volume
PV=nRT
W=-pdeltaV


The Attempt at a Solution


I started by taking difference between two pressures =2.0g/cm3 and then converted that to 2000kg/cm3. with this i used
p=m/v
v=m/p
v=407kg/2000kg/m3
v=0.2035m3 =203.5L
W=-pdeltaV
W=1.0atm(203.5L)
W= 203.5atm.L=20619 J
this does not seem correct at all and it isn't.
 
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I believe you haven't worked out ΔV correctly. Start by working out the volume at the two different temperatures. Perhaps best to convert the data given in the question to kg and kg/m3 first.
 

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