IB
HL Physics
Thermal Physics Revision Q’s
Solutions
1.
B
[1]
2.
C
[1]
3.
(a)
(b)
gas that obeys the equation pV = nRT / no forces between molecules;
at all pressures, volumes and temperatures / any other postulate;
(i)
(ii)
(c)
pV = nRT
20 × 106 × 2 × 10–2 = n × 8.3 × 290;
n = 170 (166);
2
2 max
number = n × NA;
number = 166 × 6.02 × 1023 = 1.0 × 1026;
2
(i)
average volume = 2.0 × 10–28 m3;
1
(ii)
average separation 
3
( 2.0  10 28 ) ;
= 5.8 × 10–10 m;
2
Allow solution based on sphere.
[9]
4.
B
[1]
5.
C
[1]
6.
(a)
(b)
(total) kinetic energy of molecules / particles of gas;
due to random motion of molecules;
(i)
U  q  w ;
2
1
Allow with / without delta’s, upper / lower case.
(ii)
U: constant temperature / isothermal;
w: constant volume / isochoric / isovolumetric;
q: adiabatic;
3
[6]
7.
Expansion of a gas
(a)
2.4  105 Pa;
(b)
any line through (3.0, 4.0) and (5.0, 2.4);
that is a smooth curve in correct direction;
that starts and ends on the above points;
1
3
1
(c)
(i)
(ii)
work done = area under line / curve / graph;
to get 6.1  105 J;
Accept 5.5  6.7  105 J.
2
work done would be less as adiabatic line is steeper than
isothermal line / OWTTE;
or:
no energy / heat has to be transferred to the surroundings to
maintain constant temperature / OWTTE;
1
[7]
8.
(a)
(b)
(c)
(d)
V1 V 2
;

T1 T2
V
6
T2  2 T1   300 = 900 K;
V1
2
use of
W(= pV = 12 × 105 × 4.0 × 10–3 = 4.8 × 103 J) = 4.8 kJ;
Q = (4.8 kJ + 7.2 kJ) = 12 kJ;
Award full marks for correct answer without work shown.
the (magnitude of the) temperature change is the same;
the gas is ideal and so the change in internal energy depends only
on temperature;
(i)
2
2
2
for the full cycle ∆U = 0;
therefore the net work is 4.8 – 2.6 = 2.2 kJ;
or
net thermal energy in is 12 kJ and net thermal energy out
is 7.2 + 2.6 = 9.8 kJ;
so work done is 12 – 9.8 = 2.2 kJ;
or
(ii)
work is area in loop;
area = 4.8 – 2.6 = 2.2 kJ;
2
efficiency is
2 .2
;
12
= 0.18 /18 %;
2
[10]
2