(^214) Aptitude Test Problems in Physics
As the gas goes over from state./ to state 3 (p 2 , V 2 )
(points 2 and (^3) lie on the same isotherm), the
following relations are fulfilled:
AQ (^2) = AU2+ A2,
A2-_ (PO+ PO (V2- Vo)
2
Since the final temperature of the gas in states 2
and 3 is the same, A U 1 = A U 2. In order to find
out in which process the gas receives a larger
amount of heat, we must compare the works A l
and A 3:
Al—A2= (Po+ Pi) (V2-1
(^7) 2) (Poi- P2) (V2—Vo)
2 2

(PoVi— PoV2) + (P2Vo—P2V0) < 0
2
since po V 1 < po V 2 and p 2 V 0 <pi Vo. Consequently,
A2 > A l and AQ 3 > AQI, i.e. the amount of heat
received by the gas in the process / -0- 3 is larger.
2.7. Since hydrogen diffuses through all the parti-
tions, it uniformly spreads over the entire vessel,
and in all the three parts of the vessel, the pressure
of hydrogen is
PH2= p,H2
(if a gas penetrates through a partition, its pres-
sure on both sides of the partition must be the
same in equilibrium).
Nitrogen can diffuse only through the right
partition, and hence will fill the middle and right
parts of the vessel (see Fig. 61) having the volume
(2/3)V. The pressure of nitrogen is
mN 2 3RT
P112— ttr, 72 2V •
rnH 2 RT