Download ert 108 physical chemistry assignment 1

PRT 140 PHYSICAL CHEMISTRY
ANSWER SCHEME ASSIGNMENT 1
1) A certain gas mixture is at 3450 kPa pressure and is composed of 20.0 g of O2 and 30.0 g of
CO2. Calculate the CO2 partial pressure.
Answer:
1st Step: Find the number of mole of each fraction
2nd Step: Calculate the partial pressure of CO2
2) In an industrial process, nitrogen is heated to 500K in a vessel of constant volume. If it enters
the vessel at 100 atm and 300K, Calculate the pressure that would it exert at the working
temperature if it behaved as perfect gas?
Answer:
The keyword in this question is “behave as perfect gas”. So used the equation PV = nRT
3) A perfect gas undergoes isothermal compression, which reduces its volume by 2.20 dm 3. The
final pressure and volume of the gas are 5.04 bar and 4.65 dm3, respectively. Calculate the
original of the gas in (a) bar and (b) torr.
PROGRAMME OF CHEMICAL ENGINEERING TECHNOLOGY (INDUSTRIAL CHEMICAL PROCESS)
PRT 140 PHYSICAL CHEMISTRY
ANSWER SCHEME ASSIGNMENT 1
Answer:
The PV is constant so PfVf = PiVi
This equation can be solved for either initial or final pressure, hence
Original pressure in bar:
Original pressure in torr:
4) Calculate the pressure exerted by 3 mol NO2 behaving as (Refer table in the textbook for others
information)
(a) A perfect gas law
(b) A van der Waals gas when it is confined under the following condition
i) At 300K in 34 dm3
ii) At 700K in 100 cm3
Answer:
(a) (i) as perfect gas at 300K in 34 dm3
(ii) as perfect gas at 700K in 100 cm3
(b) (i) van der Waals at 300K in 34 dm3
PROGRAMME OF CHEMICAL ENGINEERING TECHNOLOGY (INDUSTRIAL CHEMICAL PROCESS)
PRT 140 PHYSICAL CHEMISTRY
ANSWER SCHEME ASSIGNMENT 1
(i) van der Waals at 700K in 100 cm3
5) Illustrate and explain the carnot cycle diagram.
Step
∆U
qrev
wrev
AB
0
RThln V2/V1
RThln V1/V2
BC
Cv (Tc-Th)
0
Cv (Tc-Th)
CD
0
RTcln V4/V3
RTcln V3/V4
DA
Cv (Th-Tc)
0
Cv (Th-Tc)
Complete
0
R(Th-Tc)ln V2/V1
R(Th-Tc)ln V1/V2
6) A sample of 3.00 mol CH3CH2OH (g) is condensed isothermally and reversibly to a liquid at
65 oC. The standard enthalpy of methanol at 65oC is 35.3 kJ mol-1. Find the w, q, ∆U and ∆H
for this process.
Answer:
∆H = ∆condensation H - ∆vapour H
PROGRAMME OF CHEMICAL ENGINEERING TECHNOLOGY (INDUSTRIAL CHEMICAL PROCESS)
PRT 140 PHYSICAL CHEMISTRY
ANSWER SCHEME ASSIGNMENT 1
= 3 mol x 35.3 kJmol
= -105.9 kJ
Since the condensation is done isothermally and reversibly the external pressure is constant at
1 atm. Hence,
q = qp = ∆H = -106.5 kJ
W = -Pex∆V ; ∆V = Vliquid – Vvapour ≈ -Vvapour because Vliquid << Vvapour
On the assumption that methanol vapour is a perfect gas, Vvapour = nRT/P and P = Pex , since
the condensation is done reversibly reversibly. Hence,
and
7) When 178 J of energy is supplied as heat to 1.9 mol of gas molecules at constant pressure, the
temperature of the sample increased by 1.78 K. Calculate the molar heat capacities at constant
volume and constant pressure of the gas.
Answer:
8) Determine
Ho/kJ for the following reaction using the listed enthalpies of reaction:
N2H4(l) + 2 H2O2(g)
N2H4(l) + O2(g)
N2(g) + 2 H2O(l)
H2(g) + 1/2 O2(g)
H2O(l)
H2(g) + O2(g)
H2O2(l)
N2(g) + 4 H2O(l)
Ho/kJ = -622.3 kJ
Ho/kJ = -285.8 kJ
Ho/kJ = -187.8 kJ
PROGRAMME OF CHEMICAL ENGINEERING TECHNOLOGY (INDUSTRIAL CHEMICAL PROCESS)
PRT 140 PHYSICAL CHEMISTRY
ANSWER SCHEME ASSIGNMENT 1
Answer:
(1)
(2)
(3)
(2) x 2
(3) x 2
N2H4(l) + O2(g)
N2(g) + 2 H2O(l)
H2(g) + 1/2 O2(g)
H2O(l)
H2(g) + O2(g)
H2O2(l)
2H2(g) + O2(g)  2 H2O (l)
2H2(g) + 2O2(g)  2 H2O2 (l)
Ho/kJ = -622.3 kJ
Ho/kJ = -285.8 kJ
Ho/kJ = -187.8 kJ
∆ Ho/kJ = -571.6 kJ
∆ Ho/kJ = -375.6 kJ
∆H = ∆Hproduct - ∆Hreactant
= ∆H {reaction (1) + 2 x reaction (2)} – ∆H {reaction (3) x 2}
= [-622.3 kJ +(-571.6 kJ)) – (-375.6 kJ)
= -818.3 kJ
9) The combination of coke and steam produces a mixture called coal gas, which can be used as
a fuel or as a starting material for other reactions. If we assume coke can be represented by
graphite, the equation for the production of coal gas is
2 C (s) + 2 H2 (g)  CH4 (g) + CO2 (g)
Determine the standard enthalpy change for this reaction from the following standard
enthalpies of reaction:
(1) C (s) + H2O (g)  CO (g) + H2 (g)
∆Hf = 131.3kJ
(2) CO (g) + H2O (g)  CO2 (g) + H2 (g)
∆Hf = -41.2 kJ
(3) CH4 (g) +H2O (g)  3 H2 (g) + CO (g)
∆Hf = 206.1 kJ
Answer:
(1)
C(s) + H2O (g)  CO (g) + H2(g)
∆H=131.3 kJ
(2)
CO (g) + H2O(g)  CO2(g) + H2(g)
∆H=-41.2kJ
(3)
CH4(g) + H2(g)  3H2O(g) + CO(g)
∆H=206.1kJ
(1) x 2 2C(s) + 2H2O (g)  2CO (g) + 2H2(g)
∆H=262.6kJ
(2)
CO (g) + H2O(g)  CO2(g) + H2(g)
∆H=-41.2kJ
(3)
3H2O(g) + CO(g)  CH4(g) + H2(g)
∆H=-206.1kJ
2 C (s) + 2 H2 (g)  CH4 (g) + CO2 (g)
∆H=?
10.
∆H = 262.6 kJ – 41.2 kJ -206.1 kJ
= 15.3 kJ
A sample consisting of 2.0 mol of diatomic perfect gas molecules at 250 K is compressed
reversibly and adiabatically until its temperature reaches 300 K.Calculate q, w, ∆U, ∆H and
∆S. (Given that Cv,m = 27.5 JK-1mol-1).
Answer:
PROGRAMME OF CHEMICAL ENGINEERING TECHNOLOGY (INDUSTRIAL CHEMICAL PROCESS)
PRT 140 PHYSICAL CHEMISTRY
ANSWER SCHEME ASSIGNMENT 1
PROGRAMME OF CHEMICAL ENGINEERING TECHNOLOGY (INDUSTRIAL CHEMICAL PROCESS)