Quiz 3

Question 1

1st law of thermodynamics

Answer 1

Conservation of energy. It states that energy cannot be created or destroyed, only transformed or transferred. The total energy of an isolated system remains constant

Question 2

2nd law of thermodynamics

Answer 2

1. The energy in the form of heat or work can’t be extracted from a
   system unless there is a lower temperature heat reservoir available
2. Entropy tends to increase in an isolated system, meaning that energy transformations are not 100% efficient.

Question 3

entropy change of reversible thermodynamic processes is always 0, explain

Answer 3

A reversible process is a process where the system can be returned to its initial state by infinitesimally reversing the changes made, meaning there is no net increase in entropy.

Question 4

the entropy change of irreversible thermodynamic processes is always positive, explain

Answer 4

total entropy of an isolated system tend to increase, and for an irreversible process, the entropy always increases, resulting in a positive entropy change.

Question 5

The 3rd law of thermodynamics

Answer 5

The Third Law of Thermodynamics states that as the temperature of a system approaches absolute zero (0 K), the entropy of a perfect crystalline substance approaches zero. So crystalline matters are imperfect which means they can
have some entropy even at absolute zero temperature

Question 6

Thermodynamics establishes relationship between heat and what kind of energy?

Answer 6

all other forms of energy

Question 7

What is the most macroscopic property in pharmacy?

Answer 7

solubility

Question 8

environment

Answer 8

everything else outside the system

Question 9

system

Answer 9

is the part of the environment that is under study or observation

Question 10

Why is heat is an extensive property while temperature is an intensive property.

Answer 10

Heat is an extensive property because the amount of heat energy in a system depends on the quantity of matter and the conditions (e.g., a larger mass can store more heat).

Temperature, on the other hand, is an intensive property because it doesn’t depend on the size or mass of the system. Whether you have a small cup of water or a large bucket, if both are at the same temperature, this property remains the same.

Question 11

Intensive variables or properties

Answer 11

Does not depend on amount
ex. temperature, pressure, and density

Question 12

Extensive variables or properties

Answer 12

Does depend on amount
ex. mass or # of moles, volume, conc, and energy

Question 13

Adiabatic thermodynamic process

Answer 13

no energy (heat) transfer between system and environment. Temperature change as a result of work done

Question 14

Isothermal thermodynamic process

Answer 14

energy (heat) transfer can happen between system and environment. Temperature remains constant
To maintain a constant temperature, heat must be added to or removed from the system to balance the work done by or on the system

Question 15

Isobaric thermodynamic process

Answer 15

pressure remains constant, everything else changes (volume, energy, temp.)

Question 16

Isochoric thermodynamic process

Answer 16

volume remains constant, everything else changes (pressurePositive Work, energy, temp.)

Question 17

Negative Work

Answer 17

Work done by the system (energy leaving the system).
Compression of a Gas: In a process where a gas is compressed, the work done on the gas is negative because the surroundings are doing work on the gas.

Question 18

Positive Work

Answer 18

Work done on the system (energy entering the system).
Expansion of a Gas: In a thermodynamic process where a gas expands, doing work on the environment (such as lifting a piston), the work done by the gas is positive.

Question 19

Endothermic processes

Answer 19

absorbs heat e.g. melting of ice, evaporation or
vaporization of water

Question 20

Exothermic processes

Answer 20

produces heat e.g. freezing of water, combustion, etc

Question 21

Heat Capacity (extensive property):

Answer 21

Ratio of amount of heat absorbed or released and change in temperature of a system without undergoing phase transition

Question 22

Specific heat capacity

Answer 22

when amount is 1 g

Question 23

Molar heat capacity

Answer 23

when amount is 1 mole

Question 24

Latent heat

Answer 24

Heat needed to change phase happening at constant temperature and
pressure.

Question 25

A gas expands by 0.25 liter against a constant pressure of 1.5 atm at 25C. What is the work in joules done by the system?

Answer 25

W = Volume * Pressure
0.987 atm = 10^6 dynes/cm^-2 and 0.25 L = 250 cm^3
Therefore, W = 1.519 atm x 10^6 dynes/cm2 x 250 cm3 = 0.38 x 10^9 dynes/cm
So 0.38 x 10^9 dynes/cm = 0.38 x 10^9 ergs = 38.0 joules because 1 joule = 107 ergs

Question 26

What is the entropy change accompanying the vaporization of 2 moles of water in
equilibrium with its vapor at 25oC? (Heat of vaporization required to convert
water to its vapor = 10,500 cals/mole).

Answer 26

Entropy Change = Molar heat of Vaporization / Temperature
(10 ,500 cals/mole * 2 mole) / 298.15K = 70.67cals/K

Question 27

Calculate the molar heat capacity if 4000 calories of heat are required to raise the
temperature from 30 to 35oC of 4 moles a liquid.

Answer 27

Cm (molar heat capacity) = Q / n (delta T)
then, 4000cal / 4mol (35-30 = 5C)
so 4000cal / 4 mol * 5C = 200cal/mol*C

Question 28

when should Helmholtz free energy equation be used? ∆A = ∆E – T (∆S)

Answer 28

use to describe changes in state at constant temperature and volume, not constant pressure

Question 29

when should Gibbs free energy equation be used? ∆G = ∆H – T (∆S)

Answer 29

use to describe changes in state at constant temperature and pressure, not constant volume

Question 30

when should van’t Hoff equation be used

Answer 30

provides equilibrium constant at another temperature if it is known at one temperature

Question 31

∆H and ∆S for the following reaction are 28.05 kJ and 108.7 J/K respectively.

NH4NO3(s) + H2O (l) –> NH4+ (aq) + NO3-(aq)

Calculate the value of ∆G0 for this reaction at 25C and explain why this reaction will be spontaneous.

Answer 31

∆G = ∆H – T (∆S) = 28050 J – (25+273.15K) x 108.7 J/K = -4.36 J, so spontaneous because of negative ∆G value

Question 32

The concentration of urea in plasma and urine are 0.006 M and 0.345 M,
respectively. Calculate the free energy in transporting 0.01 mole of urea from plasma to urine. Is this transport process spontaneous, i.e., would it happen on its own? How many ATP molecules would be consumed in providing energy for this transport process?

Answer 32

ΔG = nRT ln(c2/c1)
N = 0.01 moles
R= 1.987 cal/moleK
T = 37 (body temp) +273.15 = 310.15 K
c2 = 0.345
c1 = 0.006
ΔG = (0.01 moles)(1.987 cal/moleK)(310.15 K) = 24.69 calories
The hydrolysis of 1 mole ATP releases 7.3 kCal (7300 calories).
So 24.69 calories * 6.0221x10^23 molecules of ATP / 7300 calories = 2.06x10^21 molecules of ATP