1st Year Thermodynamics Practice Questions Solution - Dr. Mark Gray

This document provides a comprehensive solution to a thermodynamics homework assignment. The solution covers several key concepts, including calculating heat capacity, enthalpy changes, and temperature changes in various scenarios. It addresses questions related to the heat transfer of bovine brain extract, temperature changes in the human body, and the molar enthalpy change of nitrous oxide. The assignment also explores concepts like Gibbs free energy, entropy, and the spontaneity of reactions. Detailed calculations and explanations are provided for each problem, including determining work done, energy changes, and the application of the second law of thermodynamics. The solution also includes multiple-choice questions on fundamental thermodynamic principles and provides a list of references for further study.

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Q1
Heat capacity q = Supplied heat energy
Rise∈temperature
Heat capacity q = 125
310−277
Heat capacity q = 125
33
Heat capacity q =3.7878 kJ/C
The value of q
q= mc∆ T
where m is mass c is specific heat capacity of water , q is heat energy and ∆ T is change in
absolute temperature
q= mc∆ T
q= 0.075× 4.18 ×33
q= 10.3455 kJ/mol
Value of ∆ H
At a constant pressure the formula for q is the same as the formula for ∆ H thus
∆ H =q= mc∆ T and ∆ H = 10.3455 kJ/mol
Q2
i.
From Q= nC ∆ T
Where n is the number of moles, Q is the heat flow, C is the molar heat capacity and ∆ T is the
temperature change.
Moles 54000
18 = 3000 moles

3000×75 × ∆ T = 7000000
225000 ∆ T
225000 = 7000000
225000
∆ T = 31.1 K
ii.
When there is no change in temperature
0
mC ∆ T +m ∆ vapH =7000000
m× 44000
44000 = 7000000
44000
m= 159.09g
Q3
Molar capacity at constant pressure C p ,m
∆ H =q= C p ,m ∆ T
Where C is the heat capacity ∆ T change in temperature
∆ H =38.5×(310−277)
∆ H = 1270.5 J/K
Q4
a) Um = Hm + TS ( FALSE )
b) b) Hm = Um – RT (FALSE)
c) c) Hm = Um + RT ( TRUE)
d) d) Um = Hm / TS ( TRUE)
Q5
∆ S0=∑
n
n S0 ( products )−∑
m
m S0 (reactants)

∆ S0 = {(1mole×129.2 ¿+(1 mole ×111)}- {(1mole×213.4 ¿+(1 mole × 69.9)}
∆ S0 = 240- 273.3
∆ S0 = - 33.3J/mol.K
Q6
Total energy = energy to melt ice+ energy to heat water +vaporize water
E= m.∆ Hf +mC ∆ T +m. ∆ Hv
E= ( 0.2×6 ¿ + ( 0.2× 84 ×100 ) +(0.2 × 40.7)
E= 1.2+ 1680+8.14
E= 26.14kJ
Q7
From second law of thermodynamics
∆ S= ∆ Q
T
∆ S=50000 kJ
277
∆ S=¿ 180.5054 kJ.K-1
Q8
Mass of ice = density × volume
Mass of ice = 0.8×25
Mass of ice = 20g
∆ S = Q
T =mC ∆T
T
∆ S = 20
1000 ×111.5
∆ S = 2.23 J.K-1
∆ S ≈ 2 J.K-1

Q9
i. Endothermic reaction
ii. Exothermic reaction
iii. Exothermic reaction
Q10
∆ H 0=∑ ∆ H0 f ( products ) −∑ ∆ H0 f ( Reactants )
C6H12O6+ C6H12O6 C12H22O11+H2O
The mole ratio in the reaction is 1:1:1:1
Products = {( moles×−2226 ¿+(1 mole ×−286) }=−2512
Reactant = {( moles×−1273 ¿+(1 mole ×−126 6)}=−2539
∆ H o sucrose = −2512−(−−2539)
∆ H o sucrose = 27kJ
Q11
Work done = F× d
F=mg = (65+20)× 9.8
F=833 N
Work= 833×200
W= 166.6kJ
The number of moles = 166.6
2828
The number of moles= 0.05891 moles
Mass = moles × molar mass
Mass = 0.05891 ×180
Mass = 10.6038 g

Q12
At 33ft the volume in the lungs will reduce to half the original volume. And the work done is
given by the following equation,
W=p∫Vdv
W= 200×{5.0× 1 0−4 −2.5 ×1 0−4 }
W= 200×2.5 × 10−4
W= 5 0 j
Q13
i. From the following relationship between Gibbs, enthalpy and entropy
∆ G=∆ H−T ∆ S
∆ G= -150- (310×−97 ¿
∆ G= -150+30070
∆ G= 29920J
∆ G= 29.92kJ
ii.

The reaction is not spontaneous. It is non spontaneous since the Gibbs free energy is positive.
Spontaneous reaction occurs only when the Gibbs free energy is negative.
Q14
Power = Energy
Time
Power = 10
Time = 3×60 × 60=10800 seconds
Energy = Power ×time
Energy = 10×10800
Energy = 1080000 J required for 3 hours
Number of moles = 108 kJ
2828 =0.03818 moles
Mass = moles × molar mass
Mass = 0.03818 ×180
Mass = 6.8724 grams
Q15
Never. Since any process has just one entropy change and it is either negative ( exothermic) or
positive ( endothermic).

References
Arora, C. P., 2013. Thermodynamics. 3rd ed. Hull: Tata McGraw-Hill Education.
Logan, E., 2010. Thermodynamics: Processes and Applications. 3rd ed. Liverpool: CRC Press.
Turns, S. R., 2011. Thermodynamics: Concepts and Applications. 3rd ed. Hull: Cambridge University
Press.
Vidal, J., 2010. Thermodynamics, Volume 1. 2nd ed. Hull: Editions OPHRYS.