chemistry department al-kharj, nov 2016 college of science ... · chemistry department al-kharj,...

Chemistry Department Al-kharj, Nov 2016

College of Science and Humanities Studies Exercises- Physical Chemistry I (Chem 3310)

Prince Sattam Bin Abdulaziz University

1

Exercises-Physical Chemistry I (Chem 3310)

Chapter-3- The second and third laws of thermodynamics

Exercise 1

Calculate the change in entropy when 50 kJ of energy is transferred reversibly and isothermally

as heat to a large block of copper at (a) 0°C, (b) 70°C.

Solution

We have:

𝑑𝑆 =𝑑𝑞𝑟𝑒𝑣

𝑇

For a measurable change,

∆𝑆 = ∫𝑑𝑞𝑟𝑒𝑣

𝑇

Process is reversible and isothermal (T=constant), thus

∆𝑆 =1

𝑇∫ 𝑑𝑞𝑟𝑒𝑣 =

𝑞𝑟𝑒𝑣

𝑇

qrev is the energy transferred as heat (qrev = 50 kJ).

(a) The change in entropy at T = 0°C = (0+273.15) K = 273.15 k

We have

∆S =50 × 103

273.15= 1.8 102 J k−1

(a) The change in entropy at T = 70°C = (70+273.15) K = 273.15 k

We have

∆S =50 × 103J

(70 + 273.15)= 1.5 102 J k−1




2

Exercise 2

Calculate the molar entropy of a constant-volume sample of argon at 250 K given that it is 154.84

J K−1

mol−1

at 298 K.

Data: Cp,m = 20.786 J K-1

mol-1

and R = 8.3145 J K-1

mol-1

Solution

At constant-volume, the change in molar entropy when monoatomic gas argon (Ar) is heated

from 250 K to 298 K is (Tf = 298 K and Ti = 250):

𝑆𝑚(𝑇𝑓) = 𝑆𝑚(𝑇𝑖) + 𝐶𝑉,𝑚 ∫𝑑𝑇

𝑇

𝑇𝑓

𝑇𝑖

Thus,

𝑆𝑚(𝑇𝑖) = 𝑆𝑚(𝑇𝑓) + 𝐶𝑉,𝑚 ∫𝑑𝑇

𝑇

𝑇𝑖

𝑇𝑓

= 𝑆𝑚(𝑇𝑓) + 𝐶𝑉,𝑚 ln𝑇𝑖

𝑇𝑓

where CV,m = Cp,m - R

Sm(250) = Sm(298) + CV,m ∫dT

T

298

250

= 154.84 J K−1 mol−1 + (20.786 − 8.3145) J K−1 mol−1 ln250

298

= 152.65 J K−1 mol−1




3

Exercise 3

Calculate ΔS (for the system) when the state of 2.00 mol diatomic perfect gas molecules, for

which Cp,m = (7/2)R, is changed from 25°C and 1.50 atm to 135°C and 7.00 atm. How do you

rationalize the sign of ΔS?

Solution

ΔS of the system has the same value as if the change happened by reversible heating at constant

pressure (step 1), followed by reversible isothermal compression (step 2).

ΔS = ΔS1 + ΔS2

∆𝑆1 = ∫𝑑𝑞𝑟𝑒𝑣

𝑇= 𝑛𝐶𝑝,𝑚 ∫

𝑑𝑇

𝑇

𝑇𝑓

𝑇𝑖= 𝑛𝐶𝑝,𝑚 ln

𝑇𝑓

𝑇𝑖

∆𝑆1 = (2.00 𝑚𝑜𝑙−1) × (7

2) × (8.3145 𝐽𝐾−1𝑚𝑜𝑙−1) × ln (

(135 + 273.15)𝐾

(25 + 273.15)𝐾) = 18.3 𝐽𝑘−1

For the second step (isothermal process)

∆S2 = ∫dqrev

T=

qrev

T

where

qrev = −w = ∫ pdV = nRT lnVf

Vi= nRT ln

pi

pf

Where pi = nRT/Vi and pf = nRT/Vf

Hence,

∆S2 = nRT lnpi

pf= (2.00 mol−1) × (8.3145 JK−1mol−1) ln (

1.5 atm

7.00) = −25.6 Jk−1

Thus,

ΔS = ΔS1 + ΔS2 = (18.3 – 25.6) JK-1

= -7.3 JK-1

The heat lost in step 2 was more that the heat gained in step 1, resulting in a net loss of entropy.

Alternatively, the ordering represented by confining by confining the sample to a smaller volume

in step 2 overcomes the disordering represented by the temperature rise in step 1. A negative

entropy change is allowed for a system as long as an increase in entropy elsewhere results in ΔStot

> 0.




4

Exercise 4

A sample consisting of 2.00 mol of diatomic perfect gas molecules at 250 K is compressed

reversibly and adiabatically until its temperature reaches 300 K. Given that CV,m = 27.5 J K−1

mol−1

, calculate q, w, ΔU, ΔH, and ΔS.

Solution




5

Exercise 5

Calculate ΔH and ΔStot when two iron blocks, each of mass 1.00 kg, one at 200°C and the other

at 25°C, are placed in contact in an isolated container. The specific heat capacity of iron is 0.449

J K−1

g−1

and may be assumed constant over the temperature range involved.

Solution




6

Exercise 6

The enthalpy of vaporization of methanol is 35.27 kJ mol−1

at its normal boiling point of 64.1°C.

Calculate

(a) The entropy of vaporization of methanol at this temperature

(b) The entropy change of the surroundings.

Solution




7

Exercise 7

The heat capacity of chloroform (trichloromethane, CHCl3) in the range 240 K to 330 K is given

by Cp,m /(J K−1

mol−1

) = 91.47 + 7.5 × 10−2

(T/K).

In a particular experiment, 1.00 mol CHCl3 is heated from 273 K to 300 K. Calculate the change

in molar entropy of the sample.

Solution




8




9




10




11




12




13




14

chemistry department al-kharj, nov 2016 college of science ... · chemistry department al-kharj,...

Documents