### Process calculation in ideal gas

```13/14 Semester 2
Physical Chemistry I
(TKK-2246)
Instructor: Rama Oktavian
Email: [email protected]
Office Hr.: M.13-15, Tu. 13-15, W. 13-15, Th. 13-15, F. 09-11
Outlines
1. Change in energy
2. Change in state at constant volume
3. Change in state at constant pressure
(Process calculation in ideal gas)
Change in energy
Energy is an extensive state property of the system
Energy per mole is an intensive state property of the system
Energy is conserved in all transformations
First law thermodynamic
dU = dQ + dW
dU of the system depends only on the initial and final states
Q and W depends on path
Change in energy
First law thermodynamic
dU = dQ + dW
dU of the system depends only on the initial and final states. Define U = U(T,V)
 U 
 U 
dU  
 dT  
 dV
 T V
 V T
Substituting dU from the first law thermodynamic
 U 
 U 
dQ  dW  
 dT  
 dV
  T V
 V T
Change in energy
First law thermodynamic
 U 
 U 
dQ  dW  
 dT  
 dV
  T V
 V T
dW   pdV
 U 
 U 
dQ  pdV  
 dT  
 dV
  T V
 V T
Change in energy
First law thermodynamic
 U 
 U 
dQ  dW  
 dT  
 dV
  T V
 V T
dW   pdV
 U 
 U 
dQ  pdV  
 dT  
 dV
  T V
 V T
Change in energy
First law thermodynamic
 U 
 U 
dQ  pdV  
 dT  
 dV
  T V
 V T
d ( nU )  dQ  pd ( nV )
This is the general first-law equation for a mechanically reversible, closedsystem process
Change in energy
Example of energy change – state and path function
Change in state at constant volume
First law thermodynamic
 U 
 U 
dQ  pdV  
 dT  
 dV
  T V
 V T
d ( nU )  dQ  pd ( nV )
Change state at constant volume
dV  0
 U 
dQ  
 dT
  T V
Change in state at constant volume
New properties – heat capacity
 U 
dQ  
 dT
  T V
 U 


dT
  T V
dQ
Heat capacity at constant volume
CV
 U 


  T V
Change in state at constant volume
Heat calculation at constant volume
CV
 U 


  T V
dU  C V dT
U2
 dU
U1
If Cv constant
U2
 dU
U1
T2
 CV
 dT
T1
T2

C
T1
V
dT
Change in state at constant volume
Heat calculation at constant volume
d ( nU )  dQ  pd ( nV )
dQ  dU
Valid for constant volume process
U2
Q 
 dU
U1
U2
Q  U 2 U1
 dU
T2

U1
T2
Q 
C
T1
V
dT
Heat calculation at constant volume process
C
T1
V
dT
Change in state at constant pressure
In laboratory practice most changes in state are carried out under a constant
atmospheric pressure
Change in state at constant pressure
Recall first law thermodynamics mathematical formulation
d ( nU )  dQ  pd ( nV )
Integrating this equation at constant pressure, we obtain
Change in state at constant pressure
Rearranging this equation
Change in state at constant pressure
Introducing new extensive state property of system
Enthalpy
Valid for constant-pressure process
Change in state at constant pressure
Heat calculation at constant pressure process
constant pressure process
Change in state at constant pressure
Heat calculation at constant pressure process
Heat capacity at constant pressure process
If Cp is constant
Valid for constant pressure
process
Implied property relation for ideal gas
Relation between Cv and Cp
CV
 U 


  T V
the first law statement is
dU  dW
dU   pdV
Adiabatic change state in ideal gas
C v dT   pdV
For ideal gas
pV  nRT
Integrating this equation from initial state (T1, V1) into final state (T2, V2), we have
If Cv is independent to temperature (T)
For ideal gas we have relationship
PVT relationship for adiabatic change state in gas ideal
Learning check
Check and Re-do example 7.3 from Castellan
An ideal gas, Cv = 5/2 R, is expanded adiabatically against a constant pressure of 1
atm until it doubles in volume. If the initial temperature is 25 °C, and the initial
pressure is 5 atm, calculate T2 ; then calculate Q, W, ΔU, and ΔH per mole of
gas for the transformation.
Assignment
Open your textbook (Castellan) and do these following problem:
Problem 7.1, 7.4, 7.10, 7.15, 7.17
The constant-pressure heat capacity of a sample of a perfect gas was found to vary
with temperature according to the expression Cp /(J K−1) = 20.17 + 0.4001(T/K).
Calculate q, w, ΔU, and ΔH when the temperature is raised from 0°C to 100°C (a) at
constant pressure, (b) at constant volume.
Process in ideal gas
Isothermal Process (constant temperature) for closed system process
Governing equation
Process in ideal gas
Isobaric Process (constant pressure) for closed system process
U 
C
V
Q  H
dT
Process in ideal gas
Isochoric Process (constant volume) for closed system process
Q  U
Process in ideal gas
Example
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