Bukhovtsev-et-al-Problems-in-Elementary-Physics

HEAT. MOLECULAR PHYSICS 271
324. Let us separate a cylindrical volume of the gas in direct contact
with the wall (Fig. 397). The forces acting on the side surface of the cylin..
der are mutually balanced. Since the volume is in equilibrium, the pressure
on the gas from the side of the wall should always be equal to the pressure
on the .other base of the cylinder from the side of the gas. We can conclude,
on the basis of Newton's third law, that the pressure of the gas on the wall
is equal to the pressure inside the vessel.
325. The pressure in the gas depends on the forces of interaction between
the molecules (see Problem 322). The forces of mutual interaction of the
molecules and of the molecules with the wa11 are different, however. Hence
the pressures inside the gas and at the walls of the vessel (see Problem 324)
can be identical only if the concentrations are different.
326. Since the volume is constant
Hence,
~=.~, or P2-Pl=T2- T 1 = O.004
PI T 1 PI T 1
=T 2-T1=2500K
T
1 0.004
327. From Archimedes' law, mg+G=yV, where y is the specific weight
of water and V is the volume of the sphere. The equation of state gives
(Po+ yh) V=!!!:.... RT
~
Upon deleting V from these equations, we find that
m GJ-t (Po +yh) ~ 0.666 g
VRT - fJ-g (Po + yh)
and equilibrium will be unstable.
328. When the tube is horizontal, the device cannot be used as a thermometer,
since the pressures exerted on the drop from the right and from ·the
left will be balanced at any temperature.
If the tube is placed vertically, the pressure of the gas in the lower ball
will be higher than in the ufper one by a constant magnitude. If the volume
is the same, the pressure wil grow with a rise in the temperature the faster,
the higher is the initial pressure. To maintain a constant difference of the
pressures in the balls, the drop will begin to move upward, and in this case
the device can be employed as a thermometer.
329. Since the masses of the gas are the same in both ends and the piston
is in equilibrium,
Hence,
T 2 =!.!. T 1 =330 0
VI
Applying Boyle's law to the volume of the gas whose temperature does not
change, we obtain
p= Po Vo= l. 05 atm
VI
K