Aptitude Test Problems in Physics Science for Everyone by S Krotov ( PDFDrive.com )

Solutions^217

be seen that the work is A = Mgh,

Mgh —

3

liR (Tf—Tj).

2

and hence

(3)

termwise and

we obtain the

(4)

Subtracting Eq. (1) from Eq. (2)
using expression (3) for Ti,
following equation in h:
MV 2
Mgh—pohS-- —11/Igh.
3

Hence we find that

MgV

h--

S (PoS Mg/ 3 )

Substituting h into Eq. (2), we determine the final

temperature of the gas:

Tf= (p

oS + Mg) (3p 0 S — 2Mg) V

(3poS ± Mg) SnR •

2.10. According to the first law of thermodynamics,

the amount of heat Q supplied to the gas is spent

on the change, AU in its internal energy and on

the work A done by the gas:

Q = AU+ A.

The internal energy U of a mole of an ideal gas can
be written in the form U = cv T = (3I2)RT, i.e.
AU = (3/2)R AT. The work done by the gas at
constant pressure p is A = p AV = pS Ax, where
Ax is the displacement of the piston. The gas pres-
sure is

P= Po +

Mg

s

i.e. is the sum of the atmospheric pressure and the
pressure produced by the piston. Finally, the equa-
tion of state pV = RT leads to the relation be-
tween the change AV in volume and the change A T
in temperature at a constant pressure:
p AV R AT.