{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

phys 369 09 hw1 solutions

phys 369 09 hw1 solutions - Phys 369 Fall 2009 Homework#1...

This preview shows pages 1–4. Sign up to view the full content.

Phys. 369, Fall 2009 Homework #1 Solutions 1. (5 points) 2. (8 points) The volume of the bulb is () 2 3 3 mm 4 mm 28.3 mm 4 V π == . Since we are told to assume that the volume of the cylindrical tube is negligible compared to the volume of the bulb, this is also the nominal volume of the mercury. The change in the volume of the mercury, for a change in temperature from 0 C ° to 100 C ° (a change of 100 C 100 K T Δ = ), is 41 3 3 1.81 10 K 28.3 mm 100 K 0.51 mm VV T β −− Δ= Δ= × × × = This causes the mercury to occupy an additional height 10 cm 100 mm h = = of the cylindrical tube of diameter d , of that same volume, so that 2 4 d Vh Δ= , or 1/2 1/2 3 4 4 0.51 mm 0.081 mm 100 mm V d h ππ ⎛⎞ Δ× = ⎜⎟ × ⎝⎠ 3. (10 points)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
4. (15 points) a) The number of helium atoms contained in the cylinder is 53 27 23 Pa 10 atm 1.013 10 20 m atm 4.97 10 J 1.381 10 295 K K PV N kT ×× × == (1 Pa m 3 is equal to 1 J.) The atomic weight of helium is 4.003, so the mass of one helium atom is 27 27 4.003 u 4.003 1.661 10 kg 6.65 10 kg m −− × ×= × , where u is the atomic mass constant, equal to 1/12 of the mass of one 12 C atom. Therefore the total mass of the helium is

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 7

phys 369 09 hw1 solutions - Phys 369 Fall 2009 Homework#1...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online