# Thermodynamics (density)

1. Dec 16, 2003

I'm stumped on some of the thermodynamic problems involving density.

Heat a 1.0kg bar of lead at atmospheric pressure from 25C to 60C and find the work done by the lead. (density of lead = 1.13*10^4 kg/m^3)
The only equations I was given was the Ideal Gas Law... which lead is not a gas.. Someone also told me that apparently work=pressure*volume, which also doesn't help. Density is molar mass/volume (I think), which does nothing in this case, since I basically already can determine both those components. Where do I start?

A droplet of silver has a radius of .6mm. How many silver atoms are in the droplet. The density of silver is 1.05*10^4 kg/m^3
This sounds simple, but I must be missing a conversion. I'm sure I'm suppose to multiply something by Avogradro's number, but not sure how to get the info from the density and radius.

2. Dec 16, 2003

### ShawnD

I'm thinking you would need the coefficient for linear expansion for the lead. Then you would have the change in volume. Since the pressure is N/m^2. If you multiply the pressure by the change in volume (m^3), you get Nm which is J.

find volume:
v = (4/3)$$\pi$$r^3
v = (4/3)$$\pi$$(0.0006)^3
v = 9.0478*10^-10 m^3

find mass:
(9.0478*10^-10 m^3) * (1.05*10^4 kg/m^3) = 9.5*10^-6 kg = 0.0095g

use molar mass to find mols:
0.0095g * 1mol/107.87g = 8.807*10^-5 mols

use avogadro's number to find atoms:
8.807*10^-5 mols * 6.02*10^23 atoms/mol = 5.30185*10^19 atoms

Yep, still got it