CH 353 : Quiz 1 - solution
Consider the plot below of the pressure as a function of temperature. Which curve
describes a system with a larger volume, (a) or (b)? (The number of moles is the
same). The answer
Quiz 8 Solution
Define an ideal solution.
Solution:
A solution that follows Raoults law is called an ideal solution, which means for
every component j of the solution, ! = ! ! , and ! is the vapor pressure of
Quiz 4 Solutions
1. False
In adiabatic expansion = 0, however it does not mean that = 0.
!
= !"# ,
where !"# is the heat change in reversible process. (Thus, to calculate we
need to build up reversible
Quiz 12 - Solution
Which is larger, the most probable velocity, or the average relative velocity?
S: The average relative velocity is larger.
The average relative velocity < > =
The most probable velocity !"
Quiz 2 - Solutions
True or False?
1) The parameter a in the van der Waals equation for real gases determines the
repulsion strength between gas atoms
2) In a closed system heat can be transmitted to the surrounding
Quiz 6 Solutions
Which of the thermodynamic state functions that we discussed depends on the
variables p (pressure) and V (volume)?
Solution:
None
= , = (, );
= ; = = ; = , ;
= + ; = + + = + ;
Quiz 10 Solution
Given a reversible reaction the time dependence of the relaxation of A to
equilibrium (Aeq) is
= exp
where is in terms of k1 and k-1.
Solution:
= ! + !
(
= ! + !
if at t=0, A(
Midterm key:
1 V
. We are asked to compute the
V P T
compressibility for ideal gas. From the equation of state of ideal gas we have
nRT
nRT
V
. We therefore also have
V=
= 2 . The compressibility is therefore
P T
P
P
P nRT 1
=
=
nRT P
HW 8 - solutions
Problem 1
Solution:
In this problem, we need to estimate the enthalpy of vaporization of mercury at the boiling point with the
conditions that
ln
!
!"#
=
!"#".!
!
+ 17.85, , = 3.82/, , = 12.7/.
HW 1 - solutions
1. A certain sample of gas has a volume of 0.452L measured at 87oC and 0.620 atm.
What is its volume at 1 atm and 0oC?
Solution:
= =
Thus, two states
HW3 Solutions
Problem 1.
Solution:
Reversible compression does the minimum amount of work to compress the gas.
!
!
!
!"# = =
= ln
!
!
!
!"!
= 5 8.3145 ! ! 300 ln !"! = 1.1410!
Thus, the min
HW 5 - solutions
1. One mole of ideal gas expands isothermally and irreversibly to 44.8 liters,
under conditions such that the work is -100 calories. Calculate the change in
entropy if the temperature is 300K.
HW 6 - solutions
Problem 1
Solution:
, = ! ,
( 1)
In this problem, you need to calculate !"# for this process at different temperatures.
As we know = ; = ();
For vaporization process T (temperature) is const
Midterm II : Key
dp pH
. In vaporization we convert liquid to gas
=
dT RT 2
and in sublimation solid to gas. Since H sublimation = H melting + H vaporization and all
1. Clausis-Clapeyron Eq. is
the enthalpy differences are of the same sign and pos