222 Aptitude Test Problems in Physicsthe gas remains constant, and the entire amount
of heat Q 2 supplied to the gas is spent on doing
work: Q2 = A 2.
On the isochoric segment, 3-4, the gas temper-
ature returns to its initial value T 1 , i.e. the amount
of heat Q 1 is removed from the gas. On the iso-
thermal segment 4-1, the work done by the gas
is negative, which means that some amount of
heat is taken away from the gas. Thus, the total
amount of heat supplied to the gas per cycle is
Q 1 -I- A 2. Figure 200 shows that the work done by
the gas per cycle is the sum of the positive workA (^2) on the segment 2-3 and the negative work A 4
on the segment 4-1.
Let us compare the pressures at the points
corresponding to equal volumes on the segments
4-1 and 2-3. The Gay-Lussac law indicates that
the ratio of these pressures is T 1 /T 2 , and hence the
work done by the gas is A 4 -(T 1 /T2) A 2. The
total work per cycle is given by
T i
A 2 -f -A 4 = (1— -) 2 - A 2 ,
and the efficiency is
A (^) 1— T1/T2
" Q 1 -FA2 1+Q1/A2 <1— 1'2 •
Therefore, the efficiency of the heat engine based
on the cycle consisting of two isotherms and two
isochors is lower than the efficiency 1 — Ti /T 2
of Carnot's heat engine.
2.15*. Let us first determine the free-fall accel-
eration gpi on the surface of the planet. On the one
hand, we know that the force of attraction of a
body of mass m to the planet is mgpl. On the other
hand, it follows from the law of universal gravi-
tation that this force is GznM/r 2 , where G is the
gravitational constant. Hence we obtain gpi =
GM/r 2. The pressure p exerted by the atmospheric
column of height h on the surface of the planet is
p = pgpih, (1)