09404882433 2000 00993 m 3 kg specific heats specific

) =
(0.94)(0.4882)(433)
2000
= 0.0993 m 3 /kg
Specific Heats
Specific Heat at constant volume, is defined as Cv = (
𝛿?
𝛿𝑇
) , KJ/kg K.
Cv = du/dT then’
du = Cv dT,
dU = mCv dT, Thus the first law states
δQ = dU + δW, For constant volume,
δW =
∫ 𝑝
dV = 0 , so
δQ = dU, and for constant volume process,

Q = ΔU = mCv (T
2
– T
1
), and the change in internal energy is,
ΔU = m Cv (T
2
– T
1
) = U
2
– U
1
Specific Heat at constant Pressure, is defined as , Cp = (
𝜹?
𝜹𝑻
) KJ/kg K
From h = u + pv
dh = Cp dT
dH = mCp dT
δQ = dH for P = C, hence,
δQ = dH = mCp dT, also
dh = du + R dt
Cp dT = Cv dT + R dT, thus
Cp – Cv = R
Refer to Table 5.4 , Formulas for specific heats variation with temp.
Example Problem;
For a certain ideal gas, R = 0.277 KJ/kg K and k = 1.384. What are the values of Cp and Cv?
?
Example;
Determine the average Cp value in kJ/kg K of a gas if 522 KJ of heat is necessary to raise the temperature from 300 K
to 800K making the pressure constant.

Assignments
1. A large mining company was provided with a 3000 cm 3 of compressed air tank. Air pressure in the tank drops
from 700 KPa to 180 KPa while the temperature remains unchanged at 28 0 C. What percentage has the mass
2. A mountain is to be measured by finding the change in pressure at constant temperature A barometer at the
base of the mountain reads 730 mm Hg, while at the top it reads 470 mm Hg. The local gravitational
acceleration is 9.6 m/s 2
3. For a particular gas, the value of R=0.280 KJ/kg K and the value of k=1.375. Determine the value of Cp and 4. Determine the size of spherical balloon required to lift a payload of 1360 kg. The gas to be used is Helium at
101.3 Kpa and 23
0
C. The surrounding air is 101.3 KPa and 10
0
C.
5. Helium is assumed to obey the Beattie- Bridgeman equation of state. Determine the pressure for a
temperature of 500 0 C and a specific volume of 5.2 m 3 /kg. Compare with the ideal gas equation of state. Ans.
6. Two spherical balloon each 1.81 m in diameter are connected by a pipe in which there is a valve. Each sphere
contains helium at temperature of 26.26 0 C when valve closed. One sphere contains 1.1337 kg and the other
0.56689 kg. After the valve has been open long enough for equilibrium to be obtained, what is the common
pressure in the sphere if there is no loss of energy.
7. The volume of air at 100 KPa and 25 0 C is 300m 3 . What is the volume of the same mass of air at 400 KPa and
35 0 .