(^218) Aptitude Test Problems in Physics
Substituting the expressions for AU and A
into the first law of thermodynamics and taking
into account the relation between AV and AT,
we obtain
3 5
p AV +
2
— p AV = —
2
pS Ar.
Since the amount of heat liberated by the heater
per unit time is q, Q = q At, where At is the cor-
responding time interval. The velocity of the
piston is v = Ax/At. Using Eq. (1), we obtain
2 q
v -
5 poS + Mg •
2.11. For a very strong compression of the gas,
the repulsion among gas molecules becomes sig-
nificant, and finiteness of their size should be taken
into account. This means that other conditions
being equal, the pressure of a real gas exceeds
the pressure of an ideal gas the stronger, the larger
the extent to which the gas is compressed. There-
fore, while at a constant temperature the product
pV is constant for an ideal gas, it increases with
decreasing volume for a real gas.
2.12. Let us consider an intermediate position
of the piston which has been displaced by a dis-
tance y from its initial position. Suppose that the
gas pressure is p 2 in the right part of the vessel
and pi in the left part. Since the piston is in equi-
librium, the sum of the forces acting on it is zero:
(P2 — pi) S — 2ky = 0, (1)
where S is the area of the piston.
The total work done by the gas over the next
small displacement Ay of the piston is AA
AA 1 AA 2 , where AA 2 is the work done by the
gas contained in the right part, and AA^1 is the
work done by the gas in the left part, and
AA AA 2 = p 2 AyS—pi AyS
(p 2 — Pi) AyS = 2ky Ay, (2)
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