QUIZ 1
Mata Kuliah
Tanggal
Sifat
Waktu
: Kimia Fisika
: 16 April 2014
: Closed Book
: 120 menit
1. (15 pts) Define these following terms:
a. Intensive and Extensive properties and name several intensive and extensive properties
b. Boyle’s law, Charles law, and Avogadro’s law
c. Ideal and real gas. Write all conditions when a real gas can be assumed as an ideal gas
2. a.
(20 pts) A small bubble rises from the bottom of a lake, where the temperature and pressure are
4°C and 3.0 atm, to the water's surface, where the temperature is 25°C and the pressure is 0.95 atm.
Calculate the final volume of the bubble if its initial volume was 2.1 mL
b. (20 pts) In an experiment to measure the molar mass of a gas, 250 cm3 of the gas was confined in
a glass vessel. The pressure was 152 Torr at 298 K and, after correcting for buoyancy effects, the mass
of the gas was 33.5 mg. What is the molar mass of the gas?
3. (20 pts) The gas phase reaction between nitrogen and hydrogen is as follows,
N2 + 3 H2 -----> 2 NH3
When 100 grams of N2 are added to 20 grams of H2 at 300°C ammonia is produced. If a 10 liter
reaction vessel is used, please answer the following questions.
(a) What is the pressure of the ammonia gas produced?
(b) What is the total pressure in the container AFTER reaction?
(c) What is the mole fraction of ammonia?
(d) Before reaction the pressure is higher inside the container than after reaction. Please explain why.
4. (25 pts) Thermal compressibility and volume expansivity can be expressed as function of density and
its partial derivative as follows:
 1   
1   
     and    
  p  T
   T  p
For water at 323.15 K (50°C) and 1 bar, K = 44.18 x10-6 bar-1. To what pressure must water be
compressed at 323.15 K (50°C) to change its density by 1 %? Assume that κ is independent of p.
Solution
1. a. Intensive properties – independent on the quantity of material
Ex: Pressure, temperature, specific volume, density, etc.
Extensive properties – dependent on the quantity of material
Ex: mol, volume, mass
b. Boyle’s law tells us about the relationship between the volume of a gas and its pressure at a
constant temperature, the law states that pressure is inversely proportional to the volume. In
mathematical form it can be written as
P  1/V
P x V = constant
Charles’ law tells us as the temperature of the gas increases, so does its volume, and as its
temperature decreases, so does its volume. The law says that at constant pressure, the volume of a
fixed number of particles of gas is directly proportional to the absolute (Kelvin) temperature. In
mathematical form it can be written as
VT
V = constant x T
Avogadro’s law states that the volume of a gas is directly related to the number of moles (n) of gas
at constant pressure and temperature. In mathematical form it can be written as
c. Ideal gas is gas which obeys ideal gas law for their property relationship. Ideal gas law can be
written as
PV = nRT
We can say that in ideal gas there is no attractive force and repulsive force (gas molecule
interaction is very low).
Real gas is gas which deviates from ideal gas law due to strong molecular interaction. Real gas
can be assumed as ideal gas when the pressure is very low (below 1 atm). From ideal gas law we
can see when the pressure is very low (below 1 atm), the volume of gas will be very large so that
the molecular interaction is very low.
2. a. To solve this problem we can assume that the small bubble behaves as an ideal gas. Therefore,
we can use ideal gas law relationship as follows
P1V1 P2V2
(2.1)

T1
T2
with P1 = 3 atm, T1 = 277.15 K, V1 = 2.1 mL, P2 = 0.95 atm, T2 = 298.15 K. Substituting those
values to eq (2.1), we can obtain the value of V2
V2 = 9.5 mL
b. To measure the molecular weight of ideal gas we can use eq. (2.2)
 w  RT   
M  
   RT
V  P  P 
(2.2)
From experimental data we obtain the pressure, volume, temperature and weight of gas. Based on
the data, first we need to calculate the density of observed gas.

w 3.35 x10 2 g

 1.34 x10 4 g / cm 3 , then
V
250cm 3

M    RT  16.38 g / mol
P
3. This is a limiting reactant problem.
100 gN 2
20 gH 2
 3.57 mol and
 10mol
2 g / mol
28 g / mol
H2 is limiting, therefore
3H 2 10mol

 X  3.33 mol N2
1N 2
X
The amount of ammonia produced
3H 2
10mol

 X  6.66 mol NH3
2 NH 3
X
(a) the pressure of the ammonia gas (assumed ammonia gas from reaction is gas ideal) produced
can be calculated as
nRT (6.66mol )(0.08205 atmL mol.K )(573K )

 31.31 atm
V
10 L
P
(b) 3.5714 mol N2 was available but 3.3333 mol of N2 were used leaving 0.2381 mol N2
Since it was limiting, all the H2 was consumed and 6.666 moles of NH3 were produced
Total moles = 0.2381 mol N2 + 6.666 moles of NH3 = 6.9041 mol Total
nRT (6.9041mol )(0.08205 atmL mol.K )(573K )

 32.46 atm
V
10 L
P
(c) the mole fraction of ammonia
6.666mol
 0.9655 mole fraction NH3
6.9041
(d) Look at the balanced reaction. When 1 mole of N2 and 3 moles of H2 react they become only
2 moles of NH3. So, 4 moles of reactant become 2 moles of product.
The pressure must go down.
4. This is isothermal compression process, so we will use volume expansivity equation
1   
  rearranging this equation, we will get
  p  T



2

1
 p Integrating this equation from p1 to p2 and ρ1 to ρ2 we will obtain


p2
  p with  2  1.01 2
p1
 
ln  2   p
 1 
 1.01 
p  ln 
 p  225.2 bar, p2 = 226.2 bar
  