Thermodynamics Test

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Thermodynamics Test - 10

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Weekly Quiz Competition

Question 1
1 / -0

For the process :

$$\\ \mathrm{H}_{2}\mathrm{O}(l)$$ ($$1$$ bar, $$373$$ $$\mathrm{K}$$) $$\rightleftharpoons \mathrm{H}_{2}\mathrm{O}(\mathrm{g})$$ ( $$1$$ bar, $$373 \mathrm{K}$$),

the correct set of thermodynamic parameters is

A
$$\Delta \mathrm{G}=0,\Delta \mathrm{S}=+ ve$$

B
$$\Delta \mathrm{G}=0, \Delta S=-ve$$

C
$$\Delta \mathrm{G}=+ve, \Delta \mathrm{S}=0$$

D
$$\Delta \mathrm{G}=- ve, \Delta \mathrm{S}=+ ve$$

Solution

$$\mathrm{H}_{2}\mathrm{O}(l)$$ ( 1 bar, 373 $$\mathrm{K}$$) $$\rightarrow \mathrm{H}_{2}\mathrm{O}(\mathrm{g})$$ ( 1 bar, 373 )

As this process is in equilibrium so $$\Delta \mathrm{G}=0$$

Also, here water changes from liquid to more random state gas so entropy of system increases.

$$\Delta \mathrm{S}=+$$ ve
Question 2
1 / -0

A gas mixture consists of $$2$$ moles of $$O_2$$ and $$4$$ moles of Ar at temperature $$T$$. Neglecting all vibrational modes, the total internal energy of the system is

A
11 RT

B
4 RT

C
15 RT

D
9 RT

Solution

Total internal energy $$\nu = \dfrac{f_{O_2}}{2} n_{O_2}RT+\dfrac{f_{Ar}}{2}n_{A_r}$$
$$\nu = \dfrac{5}{2}\times 2RT +\dfrac{3}{2}\times 4RT$$
$$\nu =5RT + 6RT$$
$$\nu = 11RT$$
Question 3
1 / -0

For vaporization of water at $$ 1$$ atmospheric pressure, the values of $$\Delta$$H and $$\Delta$$S are $$40.63 kJ mol^{-1} and $$108.8 JK$$^{-1}$$mol$$^{-1}$$, respectively. The temperature when Gibb's energy change ($$\Delta$$G) for this transformation will be zero, is :

A
$$273.4 K$$

B
$$393.4 K$$

C
$$373.4 K$$

D
$$293.4 K$$

Solution

According to Gibb's equation, $$\Delta$$G = $$\Delta$$H - T$$\Delta$$S when $$\Delta$$G = 0, $$\Delta$$H = T$$\Delta$$S
Given, $$\Delta$$H = 40.63 kJ mol$$^{-1}$$ = 40.63 x 10$$^3$$ J mol$$^{-1}$$
$$\Delta$$S = 108.8 J K$$^{-1}$$ mol$$^{-1}$$
$$\therefore$$ T = $$\displaystyle{\frac{\Delta H}{\Delta S} = \frac{40.63 \times 10^3}{108.8}}$$ = $$373.43 K$$
The temperature when Gibb's energy change (G) for this transformation will be zero, is $$373.4 K$$
The zero value of the Gibb's free energy change indicates equilibrium state.
Question 4
1 / -0

The following two reactions are known

$$Fe_2O_{3(s)} + 3CO_{(g)}\rightarrow 2Fe(s) + 3CO_{2(g)}$$ ; $$\Delta \ H = -26.8\ kJ$$

$$FeO_{(s)} + CO_{(g)}\rightarrow Fe_{(s)} + CO_{2(g)}\Delta H = -16.5\ kJ$$

The value of $$\Delta\ H$$ for the following reaction is:

$$Fe_2O_{3(s)} + CO_{(g)}\rightarrow 2FeO_{(s)} + CO_{2(g)}$$

A
+ 10.3 kJ

B
- 43.3 kJ

C
- 10.3 kJ

D
+ 6.2 kJ

Solution

Given :
(I) Fe$$_2$$O$$_{3(s)}$$ + 3CO$$_{(g)}$$ $$\rightarrow$$ 2Fe(s) + 3CO$$_{2(g)}$$ ; $$\Delta$$ $$H = -26.8 kJ$$

(II) FeO$$_{(s)}$$ + CO$$_{(g)}$$ $$\rightarrow$$ Fe$$_{(s)}$$ + CO$$_{2(g)}$$ ; $$\Delta$$ $$= -16.5 kJ$$

On multiplying equation (II) with 2, we get

(III) 2FeO$$_{(s)}$$ + 2CO$$_{(g)}$$ $$\rightarrow$$ 2Fe$$_{(s)}$$ + 2CO$$_{2(g)}$$ ; $$\Delta$$$$ H = -33 kJ$$

On substracting equation (IIII) from, I we get

(IV) Fe$$_2$$O$$_{3(s)}$$ + CO$$_{(g)}$$ $$\rightarrow$$ 2FeO$$_{(s)}$$ + CO$$_{2(g)}$$ ; $$\Delta$$ $$H = -26.8 - (-33) kJ = + 6.2 kJ$$

The value of $$\Delta$$H for the reaction (IV) is + $$6.2 kJ.$$
Question 5
1 / -0

For a given reaction, $$\triangle H = 35.5\ kJmol^{-1}$$ and $$\triangle S = 83.6\ JK^{-1} mol^{-1}$$. The reaction is spontaneous at: (Assume that $$\triangle H$$ and $$\triangle S$$ do not vary with temperature)

A
$$T > 298\ K$$

B
$$T < 425\ K$$

C
$$T > 425\ K$$

D
All temperatures

Solution

The reaction will be spontaneous when $$\displaystyle \Delta G$$ = negative

$$\displaystyle \Delta G = \Delta H - T \Delta S$$

$$\Delta H$$ < $$T\Delta S$$

$$\displaystyle T >\dfrac { \Delta H}{ \Delta S}$$

$$\displaystyle T = \dfrac {35.5 \: kJ/mol\times10^3\ J/kJ}{83.6 \: J/K/mol}$$

$$\displaystyle T =425$$

The reaction is spontaneous when $$\displaystyle T > 425\ K$$
Question 6
1 / -0

Consider the following processes:
$$\Delta H\left({kJ}/{mol}\right)$$
$$\dfrac{1}{2} A \rightarrow B$$ $$+150$$

$$3B \rightarrow 2C + D$$ $$-125$$

$$E + A \rightarrow 2D$$ $$+350$$

For $$B + D \rightarrow E + 2C, \Delta H$$ will be:

A
$$525\ {kJ}/{mol}$$

B
$$-175\ {kJ}/{mol}$$

C
$$-325\ {kJ}/{mol}$$

D
$$325\ {kJ}/{mol}$$

Solution

The processes are as shown below.
                                                       $$\Delta H$$
$$\displaystyle\frac{1}{2}A \longrightarrow B$$                              $$+150$$                 ............(1)
$$3B \longrightarrow 2C + D$$                     $$-125$$               ...............(2)
$$E + A \longrightarrow 2D$$                        $$+350$$                 ..................(3)
________________________
$$B + D \longrightarrow E + 2C$$     .........(4)

The reaction (1) is multiplied with 2 and added to the reaction (2). Then reaction (3) is subtracted to obtain reaction (4).

Hence, the expression for the enthalpy change for the reaction (4) is as given below.

$$\Delta H_{(4)} = 2\Delta H_{(1)}+\Delta H_{(2)}-\Delta H_{(3)}=300 - 125 - 350 = - 175$$

Hence, for $$B + D \rightarrow E + 2C,   \Delta H$$ will be $$-175\ {kJ}/{mol}$$.
Question 7
1 / -0

For the reaction, $$X_2O_4(i)\rightarrow 2XO_2(g)$$, $$\Delta U=2.1k cal, \Delta s=20 cal K^{-1}$$ at 300 K.
Hence, $$\Delta G$$ is:

A
2.7 k cal

B
-2.7 k cal

C
9.3 k cal

D
-9.3 kcal

Solution

The relationship between the enthalpy change and the change in internal energy is as shown below.

$$\Delta H=\Delta U+\Delta n_gRT$$

Substitute values in th above expression.

$$=2.1+\dfrac {2\times 2\times 300}{1000}=3.3$$

The relationship between the change in free energy, the enthalpy change and the entropy change is as shown below.

$$\Delta G=\Delta H-T\Delta s$$

$$=3.3-300\times \dfrac {26}{1000}=3.3-6=-2.7 K cal$$

Hence, the change in the free energy is -2.7 K cal.
Question 8
1 / -0

One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure, The change in internal energy of the gas during the transition is:

A
$$20\ J$$

B
$$-12\ kJ$$

C
$$20\ kJ$$

D
$$-20\ kJ$$

Solution

$$\Delta U= nC_vdT$$ whene $$n$$ is no. of moles, $$C_v$$ is heat capacity at constant volume and $$dT$$ is temperature change
for diatomic gas $$C_v=\dfrac{3}{2}R$$
from ideal gas equation $$T_1=\dfrac{P_1V_1}{R}$$ and $$T_2=\dfrac{P_2V_2}{R}$$
$$\Delta U= C_v(T_2-T_1)$$

$$\Delta U= \dfrac{3}{2}R(\dfrac{P_2V_2}{R}-\dfrac{P_1V_1}{R})$$

$$\Delta U=\dfrac{3}{2}(P_2V_2-P_1V_1)= -12\ kJ$$
hence correct answer is option B
Question 9
1 / -0

Which law of the thermodynamics helps in calculating the absolute entropies of various substances at different temperatures?

A
First law

B
Second law

C
Third law

D
Zeroth law

Solution

The third law of thermodynamics helps to calculate the absolute entropies of pure substances at different temperatures.
The entropy of the substance at temperature $$T$$ may be calculated by the measurement of heat capacity change

$$\Delta S={ S }_{ T }-{ S }_{ 0 }=\int _{ 0 }^{ T }{ \cfrac { { C }_{ P }dT}{ T } } $$

$$\Delta S={ C }_{ p }.\ln{ T } =2.303{ C }_{ p }.\log { T } $$

Note:
$${S}_{T}=$$ Entropy at $$T$$ $$K$$

$${ S }_{ 0 }=$$Entropy at $$0\ K$$
Question 10
1 / -0

Assertion (A) : Zeroth law of thermodynamics gives us the concept of energy
Reason (R) : Internal energy is dependent on temperature

A
1) Both Assertion and Reason are true and reason is correct explanation of Assertion

B
2) Both Assertion and Reason are true but reason is not correct explanation of Assertion

C
3) Assertion is true but reason is false

D
4) Assertion is false but reason is true

Solution

Zeroth law give us the definition of temperature not heat . Hence first statement is false.

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Thermodynamics Test - 10

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Question 1

$$\Delta \mathrm{G}=0,\Delta \mathrm{S}=+ ve$$

$$\Delta \mathrm{G}=0, \Delta S=-ve$$

$$\Delta \mathrm{G}=+ve, \Delta \mathrm{S}=0$$

$$\Delta \mathrm{G}=- ve, \Delta \mathrm{S}=+ ve$$

Solution

Question 2

11 RT

4 RT

15 RT

9 RT

Solution

Question 3

$$273.4 K$$

$$393.4 K$$

$$373.4 K$$

$$293.4 K$$

Solution

Question 4

+ 10.3 kJ

- 43.3 kJ

- 10.3 kJ

+ 6.2 kJ

Solution

Question 5

$$T > 298\ K$$

$$T < 425\ K$$

$$T > 425\ K$$

All temperatures

Solution

Question 6

$$525\ {kJ}/{mol}$$

$$-175\ {kJ}/{mol}$$

$$-325\ {kJ}/{mol}$$

$$325\ {kJ}/{mol}$$

Solution

Question 7

2.7 k cal

-2.7 k cal

9.3 k cal

-9.3 kcal

Solution

Question 8

$$20\ J$$

$$-12\ kJ$$

$$20\ kJ$$

$$-20\ kJ$$

Solution

Question 9

First law

Second law

Third law

Zeroth law

Solution

Question 10

1) Both Assertion and Reason are true and reason is correct explanation of Assertion

2) Both Assertion and Reason are true but reason is not correct explanation of Assertion

3) Assertion is true but reason is false

4) Assertion is false but reason is true

Solution

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