In this question paper, some of the important one mark, two and five marks questions from the chapter Thermodynamics are covered. The questions are prepared from the book back and previous year questions.

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Questions:

Multiple Choice Question

10 x 1 = 10

Mathematically, the third law of thermodynamics is expressed as _______

(a)

\(\overset{\lim}{T\rightarrow 0}{ S=0 } \)

(b)

\(\overset{\lim}{T\rightarrow \infty} {S=1 } \).

(c)

ΔU = q + w

(d)

G = H - TS

A portion of matter under consideration, which is separated from rest of universe by real or imaginary boundaries is called_____________

(a)

surroundings

(b)

system

(c)

boundary

(d)

Universe

A fundamental goal of thermodynamics is the prediction of __________ of the process.

(a)

reversibility

(b)

rate

(c)

spontaneity

(d)

none of these

In an isothermal reversible compression of an ideal gas the sign of q, ΔS and w are respectively _______________

(a)

+, -, -

(b)

-, +, -

(c)

+, -, +

(d)

-, -, +

The temperature of the system, decreases in an _____________

(a)

Isothermal expansion

(b)

Isothermal Compression

(c)

adiabatic expansion

(d)

adiabatic compression

When 15.68 litres of a gas mixture of methane and propane are fully combusted at 0° C and 1 atmosphere, 32 litres of oxygen at the same temperature and pressure are consumed. The amount of heat of released from this combustion in KJ is _____________ (ΔH_c (CH₄) = - 890 KJ mol^-1and ΔH_c(C₃H₈ = - 2220 KJ mol^-1)

(a)

- 889 kJ mol^-1

(b)

- 1390 kJ mol^-1

(c)

- 3180 kJ mol^-1

(d)

- 635.47 kJ mol^-1

Given that C_(g)+ O_2(g)⟶ CO_2(g)ΔH^o =-akJ; 2CO_(g)+O_2(g) ⟶ 2CO_2(g)ΔH^o= -bkJ; Calculate the AHo for the reaction C_(g)+ 1/2O_2(g)⟶ CO_(g) ______________

(a)

\(\frac{b+2a}{2}\)

(b)

2a-b

(c)

\(\frac{2a-b}{2}\)

(d)

\(\frac{b-2a}{2}\)

The value of ΔH for cooling 2 moles of an ideal monatomic gas from 125° C to 25° C at constant pressure will be [given C_p= \(\frac{5}{2}\)R] ____________.

(a)

- 250 R

(b)

- 500 R

(c)

500 R

(d)

+ 250 R

Which of the following is not a thermodynamic function ?

(a)

internal energy

(b)

enthalpy

(c)

entropy

(d)

frictional energy

10.

The enthalpies of formation of Al₂O₃ and Cr₂O₃are -1596 kJ and -1134 kJ, respectively. ΔH for the reaction 2Al + Cr₂O₃ ⟶ 2Cr + Al₂O₃ is _______________

(a)

- 1365 kJ

(b)

2730 kJ

(c)

- 2730 kJ

(d)

- 462 kJ

Generic 2

6 x 2 = 12

11.

What is the aim of the study of chemical thermodynamics?

12.

Two litres of an ideal gas at a pressure of 10 atm expands isothermally into vacuum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion.

13.

Give expressions for the entropy change a-phase change.

14.

Explain why C_p is always greater than C_v?

15.

Define molar heat capacity at constant volume and molar heat capacity at constant pressure.

16.

Give examples for extensive and intensive properties.

Generic 3

6 x 3 = 18

17.

Given N₂(g) + 3H₂(g) \(\rightarrow\) 2NH₃(g); \(\Delta \)H°=-92.4 kJ mol^-1. What is the standard enthalpy of formation of NH₃(g)?

18.

Enthalpies of formation of CO(g),CO₂(g), N₂O(g) and N₂O₄(g) are -110, -393, + 81 and + 9.7 kJ mol respectively. Find the value of \(\Delta\)H for the reaction. N₂O₄(g) + 3CO(g) \(\rightarrow\) N₂O(g) + 3CO₂(g).

19.

1g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298K and 1 atm pressure according to the equation.
C(graphite) + O2_(g) \(\rightarrow\) CO₂(g)
During the reaction, temperature rises. from 298K to 299K. If the heat capacity of the bomb calorimeter is 20.7 kJ mol^-1. What is the enthalpy change for the above reaction at 298K and 1 atm?

20.

Derive an expression for the efficiency of a heat engine.

21.

Show that \(\Delta\)U = nC_v (T₂ - T₁) and \(\Delta\)H = nC_p(T₂ - T₁).

22.

Define the relationship C_p - C_v = nR for an ideal gas.

Generic 5

1 x 5 = 5

23.

Calculate ΔG⁰ for the reaction. CO(g)+ \(\\ \frac { 1 }{ 2 } \)O₂(g) ⟶ CO₂(g), ΔH⁰ = - 282.84 kJ Given, S⁰(C0₂) =213.8Jk^-1 mol^-1, S^o (CO) = 197.9 Jk^-1 mol^-1 S⁰(O₂) = 205.0 Jk^-1 mol^-1

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Answers Key:

Multiple Choice Question

10 x 1 = 10

(a)
\(\overset{\lim}{T\rightarrow 0}{ S=0 } \)

(b)
system

(d)
-, -, +

(d)
- 635.47 kJ mol^-1

(d)
\(\frac{b-2a}{2}\)

(b)
- 500 R

(d)
frictional energy

10.

(d)
- 462 kJ

Generic 2

6 x 2 = 12

11.

The main aim of the study of chemical thermodynamics is to learn,
(i) Transformation of energy from one form into another form.
(ii) Utilization of various forms of energies.
(iii) Change in the properties of system produced by chemical or physical effects.

12.

Work done in free expansion or work done against zero pressure is also zero.
w_exp= -P_ext (\(\Delta\)U)
0 = w_exp=-P_ext (V₂-V₁)
0 = 0(10-2)
=0
Hence heat change, and work done is zero.

13.

\(\Delta S=\frac { { q }_{ rev } }{ T } =\frac { \Delta { H }_{ rev } }{ T } \)
Where \(\Delta\)H_rev is the enthalpy change at Temperature (T)

14.

At constant pressure processes, a system has to do work against the surroundings. Hence, the system would require more heat to effect a given temperature rise than at constant volume, so C_p is always greater than C_v.

15.

(i) The heat capacity at constant volume (C_v) is defined as the rate of change of internal energy with respect to temperature at constant volume.
\({ C }_{ v }={ \left( \frac { \partial H }{ \partial T } \right) }_{ v }\)
(ii) The molar heat capacity at constant pressure (C_p ) can be defined as the rate of change of enthalpy with respect to temperature at constant pressure.
\({ C }_{ P }={ \left( \frac { \partial H }{ \partial T } \right) }_{ P }\)

16.

Extensive properties	Intensive properties
volume, mass, amount of substance. (mole), energy, enthalpy, entropy, free energy, heat capacity.	molar volume, density, molar mass, molarity mole fraction, molality, specific heat capacity.

Generic 3

6 x 3 = 18

17.

The given reaction is N₂(g) + 3H₂(g) \(\rightarrow\) 2NH₃(g); \(\Delta\)H° = -92.4 kJ mol^-1
\(\Delta { H }^{ ° }=2\Delta { H }_{ f }^{ o }({ NH }_{ 3 })-[\Delta { H }_{ f }^{ o }({ N }_{ 2 })+3\Delta { H }_{ f }^{ o }({ H }_{ 2 })]\)
By definition, the standard enthalpy of formation of elements is equal to zero.
-92.4 = 2\(\Delta \)\({ H }_{ f }^{ o }\)(NH₃)-(0+3\(\times\)0)
i.e., for 2 mol of NH₃ 2\(\Delta\)\({ H }_{ f }^{ o }\)(NH₃)-90+3\(\times\)0)
i.e., for 2 mol of NH₃ 2 \(\Delta\)\({ H }_{ f }^{ o }\)(NH₃) = -92.4
for 1 mol of NH₃ \(\Delta\)\({ H }_{ f }^{ o }\) (NH₃) = \(\frac { -92.4 }{ 2 } =46.2\) kJ mol^-1

18.

\(\Delta { H }_{ f }^{ o }({ N }_{ 2 }{ O }_{ 4 })=+9.7\quad J\quad mol^{ -1 }\)
\(\Delta { H }_{ f }^{ O }(CO)=-110\quad kJ\quad mol^{ -1 }\)
\(\Delta { H }_{ f }^{ o }({ N }_{ 2 }O)=+81.0\quad kJ\quad { mol }^{ -1 }\)
\(\Delta { H }_{ f }^{ O }({ CO }_{ 2 })=-393\quad kJ\quad { mol }^{ -1 }\)
\(\Delta H°=\sum { \Delta { H }_{ f }^{ o } } (products)-\sum { \Delta { H }_{ f }^{ o } } (reactants)\)
\(=\left[ \Delta { H }_{ f }^{ o }({ N }_{ 2 }O)+3\Delta { H }_{ f }^{ o }{ CO }_{ 2 } \right] -\left[ \Delta { H }_{ f }^{ o }({ N }_{ 2 }{ O }_{ 4 })-3\Delta { H }_{ f }^{ 0 }(CO) \right] \)
= [81+3(-393)]-[9.7+3(-110)]
=-777.7 KJ

19.

Heat evolved in the reaction (\(\Delta\)U) = -C_v\(\Delta\)T
= -20.7 \(\times\) 1 = -20.7 kJ
Heat evolved in the combustion of 1 mol of carbon i.e., 12g of graphite
= -20.7 \(\times\)12 kJ mol^-1
= -2.48 \(\times\) 10² kJ mol^-1
In the given reactions \(\Delta\)n = 0
\(\Delta\)H = \(\Delta\)U = -2.48\(\times\)10²kJ mol^-1

20.

Efficiency = work performed / heat absorbed
\(\eta =\frac { |{ q }_{ h }|-|{ q }_{ c }| }{ |{ q }_{ h }| } \)
q_h - heat absorbed from the hot reservoir
q_c - heat transferred to cold reservoir
\(\eta =1-\frac { |{ q }_{ c }| }{ |{ q }_{ h }| } \) ....(1)
For a reversible cyclic process
\(\Delta\)S(universe) = \(\Delta\)S(system)+\(\Delta\)S(surroundings) = 0
\(\Delta\)S(system) = -\(\Delta\)S(surroundings)
\(\frac { qh }{ { T }_{ h } } =\frac { -{ q }_{ c } }{ { T }_{ c } } \)
\(\frac { -{ T }_{ c } }{ { T }_{ h } } =\frac { { q }_{ c } }{ { q }_{ h } } \)
\(\frac { { T }_{ c } }{ { T }_{ h } } =\frac { |{ q }_{ c }| }{ |{ q }_{ h }| } \) .....(2)
Substituting (2) in (1)
\(\Rightarrow \eta =1-\frac { { T }_{ c } }{ { T }_{ h } }...(3)\)
T_h >> T_c
Hence, \(\eta \) < 1

21.

For one mole of an ideal gas, we have
\({ C }_{ v }=\frac { dU }{ dT } \)
dU=C_v dT
For a finite change, we have
\(\Delta U={ C }_{ v }\Delta T\)
\(\Delta U={ C }_{ v }({ T }_{ 2 }-{ T }_{ 1 })\)
and for n moles of an ideal gas we get \(\Delta U=n{ C }_{ v }({ T }_{ 2 }-{ T }_{ 1 })\) .....(1)
Similarly for n moles of an ideal gas we get \(\Delta H=n{ C }_{ p }({ T }_{ 2 }-{ T }_{ 1 })....(2)\)

22.

The enthalpy of a system is given by
H = U + PV .....(1)
for 1 mole of an ideal gas
PV = nRT ......(2)
By subsitituting (2) in (1) i.e. PV in equation (1) by nRT
H = U + nRT ......(3)
Differentiating the above equation with respect to T,
\(\frac { \partial H }{ \partial T } =\frac { \partial U }{ \partial T } +nR\frac { \partial T }{ \partial T } \)
C_p = C_v + nR .....(1)
C_p- C_v+nR .....(4) \(\begin{bmatrix} \because { \left( \frac { \partial H }{ \partial T } \right) }_{ p } & ={ C }_{ p } \\ and\quad { \left( \frac { \partial U }{ \partial T } \right) }_{ v } & ={ C }_{ v } \end{bmatrix}\)

Generic 5

1 x 5 = 5

23.

ΔS⁰ = ΣS⁰(products) - ΣS⁰(reactants)
= [S⁰(CO₂) - S⁰(CO) + \(\\ \frac { 1 }{ 2 } \)S⁰(O₂)]
= 213.8 - [197.9 + \(\\ \frac { 1 }{ 2 } \)ｘ205]
= - 86.6 Jk^-1 mol^-1
we know, ΔG⁰ = ΔH⁰ - FΔS⁰
= - 282.84 - 298 x (- 86.6 x 10^-3)
= - 282.84 + 25.807
= - 257.033 kJ