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Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
 The specific heat capacity is defined as the
quantity of heat required to raise the
temperature of 1 kg of the gas by 10 K.
 Experiment shows that the temperature rise
of liquid water due to heat transfer to
 the water is given by
 Q = m c (T2 - T1) --------------- (8)
Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
 Where
Q = heat transfer to the water, m = mass of
water, T2 - T1 = temperature rise of the water, c
= specific heat
 In general, the value of specific heat c
depends on the substance in the system, the
change of state involved, and the particular
state of the system at the time of transferring
heat.
 Specific heat of solids and liquids is only a
function of temperature but specific heat of
Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
 Specific heat at constant volume is the
change of specific internal energy with
respect to temperature when the volume is
held constant (Isochoric process)
 -----------
--(9)
 For constant volume process:
----
---------(10)Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
 Specific heat at constant pressure is the
change of specific enthalpy with respect to
temperature when the pressure is held
constant (Isobaric process).
----
---------(11)
 For constant pressure process
--
 Specific heat at constant volume represents the
heat supplied to a unit mass of the system to raise
its temperature through 10K, keeping the volume
constant.
 Since, V= Constant, dV=0 and the work done by the
system W = PdV = 0.
 The first law of thermodynamics says:
 dQ = (dU+dW) = (dU+PdV) = dU.
 Specific heat at constant pressure represents the
heat supplied to a unit mass of the system to raise
its temperature through 10K, keeping the pressure
constant.
Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
 The first law of thermodynamics says: Q = (dU+W)
= (dU+PdV) > dU.
 As can be seen from the above, we need a quantity
equal to dU units of heat to raise the temperature
by 10K under constant volume conditions, where as
we need a greater quantity, (dU + W)>dU units of
heat to raise the temperature by 10K under
constant pressure conditions.
 Thus we find that we need more heat to perform
the work (as PdV≠0) along with to raise the
temperature of unit mass of the system through
10K under constant pressure conditions, compared
to the heat required to only raise the temperature
Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
 Let us consider one mole of an ideal gas
enclosed in a cylinder provided with a
frictionless piston of area A.
 Let P, V and T be the pressure, volume and
absolute temperature of gas respectively.
 dQ = dU + PdV
Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
 For Isochoric Change
 A quantity of heat dQ is supplied to the gas. To
keep the volume of the gas
 constant, a small weight is placed over the piston.
The pressure and the temperature of the gas
increase to P + dP and T + dT respectively.
 This heat energy dQ is used to increase the
internal energy dU of the gas. If heat dQ is
absorbed at constant volume i.e. Isochoric
 change, as dV = 0 so gas does not do any work (dW
= 0)
 ∴ PdV = 0 and dQ = CvdT for one mole of a gas
 Now, So equation 1st law becomes
Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
 For Isobaric change
 The additional weight is now removed from the
piston. The piston now moves upwards through
a distance dx, such that the pressure of the
enclosed gas is equal to the atmospheric
pressure P.
 The temperature of the gas decreases due to
the expansion of the gas.
 Now a quantity of heat dQ’ is supplied to the
gas till its temperature becomes T + dT.
 This heat energy is not only used to increase
the internal energy dU of the gas but also to do
external work dW in moving the piston
upwards.Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
 Now, heat dQ is absorbed at constant pressure,
 Then dQ = CpdT --------------- (14)
 Put value of dU and dQ from equation 13 and 14
resp. in equation 12
 CpdT = CvdT+ PdV
 (Cp – Cv) dT = PdV --------------- (15)
 For 1 mole of perfect gas, PV=RT differentiating
this equation we get
 PdV + VdP = RdT
 But for isobaric process VdP=0 above equation
becomes
 PdV = RdT --------------- (16)
 Putting the value of PdV in equation 1.5 we get
 (Cp – Cv) dT = RdT --------------- (17)Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
 Here, Cp and Cv are molar specific heat
capacities of an ideal gas at constant
 pressure and volume and R is the universal
gas constant.
 The ratio of Cp and Cv is denoted by γ
𝛾 = Cp/Cv -------------(18)
Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
 A curve showing variation of volume of a
substance taken along the X-axis and the
variation of pressure taken along Y-axis
is called an indicator diagram or P-V
diagram.
 The shape of the indicator diagram shall
depend on the nature of the
Thermodynamical process the system
undergoes.
 Let us consider one mole of an ideal gas
enclosed in a cylinder fitted with aDr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
 If dV is an infinitesimally small increase in
volume of the gas during which the pressure P
is assumed to be constant, then small amount
of workdone by the gas is dW = PdV
In the indicator diagram dW = area under the
curve a1b1c1d1.
 ∴ The total workdone by the gas during
expansion from V1 to V2 is Which is equal to
area ABCD, in the indicator diagram.
 Total workdone = 𝑉2
𝑉1
𝑃 𝑑𝑉 =Area ABCD
 Hence, in an indicator diagram the area
under the curve represents the work
done as shown in figure.
Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,

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Thermodynamic basics-3

  • 1.
  • 2.
  • 3. Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
  • 4.  The specific heat capacity is defined as the quantity of heat required to raise the temperature of 1 kg of the gas by 10 K.  Experiment shows that the temperature rise of liquid water due to heat transfer to  the water is given by  Q = m c (T2 - T1) --------------- (8) Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
  • 5.  Where Q = heat transfer to the water, m = mass of water, T2 - T1 = temperature rise of the water, c = specific heat  In general, the value of specific heat c depends on the substance in the system, the change of state involved, and the particular state of the system at the time of transferring heat.  Specific heat of solids and liquids is only a function of temperature but specific heat of Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
  • 6.  Specific heat at constant volume is the change of specific internal energy with respect to temperature when the volume is held constant (Isochoric process)  ----------- --(9)  For constant volume process: ---- ---------(10)Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
  • 7.  Specific heat at constant pressure is the change of specific enthalpy with respect to temperature when the pressure is held constant (Isobaric process). ---- ---------(11)  For constant pressure process --
  • 8.  Specific heat at constant volume represents the heat supplied to a unit mass of the system to raise its temperature through 10K, keeping the volume constant.  Since, V= Constant, dV=0 and the work done by the system W = PdV = 0.  The first law of thermodynamics says:  dQ = (dU+dW) = (dU+PdV) = dU.  Specific heat at constant pressure represents the heat supplied to a unit mass of the system to raise its temperature through 10K, keeping the pressure constant. Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
  • 9.  The first law of thermodynamics says: Q = (dU+W) = (dU+PdV) > dU.  As can be seen from the above, we need a quantity equal to dU units of heat to raise the temperature by 10K under constant volume conditions, where as we need a greater quantity, (dU + W)>dU units of heat to raise the temperature by 10K under constant pressure conditions.  Thus we find that we need more heat to perform the work (as PdV≠0) along with to raise the temperature of unit mass of the system through 10K under constant pressure conditions, compared to the heat required to only raise the temperature Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
  • 10.  Let us consider one mole of an ideal gas enclosed in a cylinder provided with a frictionless piston of area A.  Let P, V and T be the pressure, volume and absolute temperature of gas respectively.  dQ = dU + PdV Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
  • 11.  For Isochoric Change  A quantity of heat dQ is supplied to the gas. To keep the volume of the gas  constant, a small weight is placed over the piston. The pressure and the temperature of the gas increase to P + dP and T + dT respectively.  This heat energy dQ is used to increase the internal energy dU of the gas. If heat dQ is absorbed at constant volume i.e. Isochoric  change, as dV = 0 so gas does not do any work (dW = 0)  ∴ PdV = 0 and dQ = CvdT for one mole of a gas  Now, So equation 1st law becomes Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
  • 12.  For Isobaric change  The additional weight is now removed from the piston. The piston now moves upwards through a distance dx, such that the pressure of the enclosed gas is equal to the atmospheric pressure P.  The temperature of the gas decreases due to the expansion of the gas.  Now a quantity of heat dQ’ is supplied to the gas till its temperature becomes T + dT.  This heat energy is not only used to increase the internal energy dU of the gas but also to do external work dW in moving the piston upwards.Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
  • 13.  Now, heat dQ is absorbed at constant pressure,  Then dQ = CpdT --------------- (14)  Put value of dU and dQ from equation 13 and 14 resp. in equation 12  CpdT = CvdT+ PdV  (Cp – Cv) dT = PdV --------------- (15)  For 1 mole of perfect gas, PV=RT differentiating this equation we get  PdV + VdP = RdT  But for isobaric process VdP=0 above equation becomes  PdV = RdT --------------- (16)  Putting the value of PdV in equation 1.5 we get  (Cp – Cv) dT = RdT --------------- (17)Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
  • 14.  Here, Cp and Cv are molar specific heat capacities of an ideal gas at constant  pressure and volume and R is the universal gas constant.  The ratio of Cp and Cv is denoted by γ 𝛾 = Cp/Cv -------------(18) Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
  • 15.  A curve showing variation of volume of a substance taken along the X-axis and the variation of pressure taken along Y-axis is called an indicator diagram or P-V diagram.  The shape of the indicator diagram shall depend on the nature of the Thermodynamical process the system undergoes.  Let us consider one mole of an ideal gas enclosed in a cylinder fitted with aDr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
  • 16.  If dV is an infinitesimally small increase in volume of the gas during which the pressure P is assumed to be constant, then small amount of workdone by the gas is dW = PdV In the indicator diagram dW = area under the curve a1b1c1d1.  ∴ The total workdone by the gas during expansion from V1 to V2 is Which is equal to area ABCD, in the indicator diagram.  Total workdone = 𝑉2 𝑉1 𝑃 𝑑𝑉 =Area ABCD  Hence, in an indicator diagram the area under the curve represents the work done as shown in figure. Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
  • 17. Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,
  • 18. Dr. Mrs. Pritee M. Raotole, MGSM’s Arts Science and Commerce,