to the heat transferred per second per unit area of cross-section
when unit temperature gradient is set up normal to the area. k
= Area x temperature gradient Rate x heat transferred PHY 113
cxxviii k = A ( - ) Q/t 2 1 . (7.4) L
UNIT of k: From Eq.(7.4),
reduces the volume V of the gas to (V - ). These corrections
therefore enable us to re-express PV = nRT as: (P + ) (V - ) =
nRT . (5.31) R = PHY 113 cviii We
therefore need to find suitable expressions for and . It was
Van der Waal (1910) a Dutch Professo
be flat and clean. Some Vaseline may be smeared over it in
order to improve thermal contact (Fig. 7.10). Heat therefore
flows from chamber A through the specimen to the chamber B.
Steam is passed through chamber C until a steady state in
temperatures 1 an
to calculate the number of moles n numerically, divide the mass
m in grammes by the molecular mass M. n = M m m = nM
. (6.2) PHY 113 cxv
Therefore the Eq. (6.1), becomes: Q =
nMc. (6.3) Mc n Q =
. (6.4) The product Mc is called the
molar heat capacity C
of transfer of heat t Q is defined as t Q = k1A 1 2 1 d ( - ) for
air where, k1 = thermal conductivity of air For the brick also, t Q
= k2A 2 2 1 d ( - ) where, k2 = thermal conductivity of the
brick Since t Q is the same for the two materials. k1A 1 2 1
given mass of gas is directly proportional to its absolute
temperature provided the pressure is kept constant another
form of Charless law. PHY 113 xcvii SELF ASSESSMENT EXERCISE
2 Some hydrogen gas a volume of 200cm3 at 15o C. If the
pressure remains con
them in the steady state when the outside temperatures of
brick are 60o C and 10o C respectively and the area of crosssection of each is 2m2 ? 3.4 Measuring the Thermal
Conductivity of a Good Conductor It is possible to determine
the thermal conductivity
mass of hydrogen is 2g therefore 1kg = 1000g which is 500
times the mass of 1 mole. CV = 500 CV for hydrogen
Remember the unit of CV is J kg-1 K-1 or J kg-1o C-1 . At
constant volume, therefore, all the heat supplied to 1 mole of
the gas is used in raisin
the Eq. (5.27) PV = KT where K = 273 PoVo = T P V o o o The
pressure Po and 273 (To) fix the density of the gas when the
volume Vo is proportional to the mass of the gas considered.
Therefore K varies directly as the mass of the gas K is constant
in the s
done in separating molecules against attractive forces between
them is ignored. This is not so with real gases where the Van
der Waals forces have to be considered. It is also assumed that
there is no frictional force when the piston moves. 3.2 Molar
Heat
steady state condition The rate of conduction of heat, the
area of conduction, the thermal conductivity as well as the
temperature are related as t Q = kA L ( - ) 2 1 This
relationship was used to consider the temperature gradient
when a conductor is lagg
produce the equation of state for an ideal gas. At constant
pressure, PHY 113 civ PV = PoVo(1 + rt) .
. (5.22) At constant pressure, PV = PoVo(1 + t)
. (5.23) But r = = 0.0036608 = 273 1
(nearly) Thus PV = PoVo + t 273 1 1 .
(5.24) = PoVo + 273 273 t wher
the actual temperature using the Celsius scale and not for any
selected temperature rise. The valve of r for most gases is
273 1 . Now substtutng the value of r in Eq. (5.9), we get
Vt = Vo + 273 273 t . 5.10)
But as you know from the absolute scale, (2
T is inserted to measure the temperature of the bath. The initial
volume V1 and the initial temperature t1 of the gas in the bulb
are measured. The temperature of the gas is the same as that of
the water in the bath. They are both recorded when the level
conditions: Lagged material when the material is lagged
Unlagged material when the material is unlagged We shall
now describe the conduction of heat through two bars one
lagged and the other unlagged. In this section we shall consider
the conduction of h
SUMMARY In this unit, you have learnt about: The gas laws
such as: (i) Boyles law (PV = constant) (ii) Charless law ( T V =
constant) PHY 113 cix (iii) Pressure law ( T P = constant) The
three laws were combined to form the equation of state for
ideal gas
Flowers, B.H. and Mendoza, E (1970). Properties of Matter,
London: John Wiley and Sons Limited. Noaks, M.A. (1963). New
Intermediate Physics, Canada: Macmillan and company Limited.
Soars, F.W., Zemansky, M.W. and Young, H.D. (1980). College
Physics. Londo
graph of 2 - 1 against x to produce the line AB. (Fig. 7.2). In
simple words, temperature gradient, t Q , flowing through every
cross-section from the hot to cold end is constant as no heat
passes through the sides. Fig. 7.3: Temperature Gradient for a
Un
both the land and the water. But as a result of the difference in
the specific capacities of the land and water, the land is hotter
than the water. Thus warm air rises up and its place taken up by
the colder sea breeze (Fig. 8.4(i). During the night the l
they float according to Archimedes Principle. The heavier
colder molecules then take the places of the hot ones and so a
current is created within the bulk of the water. This is the
principle applied in the domestic hot water system in cold
countries. The
pressure at 0o Cfor each Celsius degree rise in temperature.
Whereas, T P = constant is the deduction or the consequence
from the law. It states that the pressure of a given mass of gas is
directly proportional to its absolute temperature. SELF
ASSESSMENT
Given that the thermal conductivity of copper is 400Wm-1o C-1
and that of the felt is 0.04Wm-1o C-1 . At a steady state
condition, the rate of heat transfer t Q is the same for copper
(AB) and the felt (BC). Using Eq. (7.4), the temperature gradient
g is
conduction take place in fluids (liquids and gases)? Thus, in this
unit, we would show whether conduction takes place in fluids
or not. This, therefore, brings us to the study of another mode
of transfer of heat energy through matter described as
CONVECTI
quantity of heat absorbed is to be determined with gases. The
molar heat capacity of a gas is considered under constant
volume or under constant pressure. When an amount of gas (n)
absorbs heat under constant volume we use the molar heat
capacity at const
London: Addison Wesley Publishing Company. Tyler, F. (1983).
Heat and Thermodynamics, London: Edward Arnold Publisher
Limited. Wilkinson, J. (1983). Essentials of Physics, Australia:
McMillan Education Publishers. Zemansky, M. W., (1968). Heat
and Thermod
1.0129 x 105 Nm-2 PHY 113 cvii For a standard temperature T =
273K and standard volume V = 22.4 litres or 22.4 x 103 m3 and
for one mole of the gas, R can be obtained by suing the Eq.
(5.28). The value of R PV = nRT R = nT PV where n = 1 1.0129 x
105 -10
(7.13) Given the area A and the thickness L of the
cardboard, K can be determined. K = mC b a x ( ) L 2 1 x A 1
(7.14) 4.0 CONCLUSION In this unit we
have studied one mode by which heat is transferred through
matter-conduction. Conduction of heat throug
EXERCISE 1 Under a Pressure of 14Nm-2 , some air has a
volume of 1.5m3 . Determine its volume when its pressure is
10Nm-2 . Assuming the temperature is kept constant. 3.3
Charless Law Charless law deals with the behaviour of a given
mass of gas at constan
and Archimedes Principle to explain convection in fluids.
Newtons law of cooling was used to quantify the rate of loss of
heat under natural or forced condition, which is proportional to
the excess temperature of its temperature and its surrounding
temper