Thermodynamics Test

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Thermodynamics Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

A spontaneous process may be defined as :

A
A process which is exothermic and evolves a lot of heat

B
A process which is slow and reversible

C
A process which takes place only in presence of a catalyst

D
A process that occurs without any input from the surroundings
Question 2
1 / -0

Ideal gas is contained in a thermally insulated and rigid container and it is heated through a resistance $$100\Omega $$ by passing a current of $$1A$$ for five minutes, then change in internal energy of the gas is

A
$$0kJ$$

B
$$30kJ$$

C
$$10kJ$$

D
$$20kJ$$

Solution

$$\Delta W=0$$
$$\therefore \Delta Q=\Delta U$$
$$\Delta Q=\Delta U={ I }^{ 2 }R\Delta t={ 1 }^{ 2 }(100)(5\times 60)=30kJ$$
Question 3
1 / -0

Directions For Questions

The electrochemical cell shown below is a concentration cell.
$$M | M^{2+} \left (\text {Saturated solution of sparingly soluble salt}, MX_{2}\right ) | | M^{2+}(0.001\ mol\ dm^{-3})|M$$
The emf of the cell depends on the difference in concentrations of $$M^{2+}$$ ions at the two electrodes. The emf of the cell at $$298\ K$$ is $$0.059\ V$$.

...view full instructions

The value of $$\triangle G (kJ\ mol^{-1})$$ for the given cell is: (take $$1F = 96500\ C\ mol^{-1})$$

A
$$-5.7$$

B
$$5.7$$

C
$$11.4$$

D
$$-11.4$$

Solution

$$\triangle G = -nFE$$
$$= -2\times 96500\times 0.059$$
$$= -11.4\ kJ/mol$$.
Question 4
1 / -0

A piece of ice kept at room temperature melts of its own. This reaction is governed by which law?

A
Second law of thermodynamics

B
First law of thermodynamics

C
Third law of thermodynamics

D
Zeroth law of thermodynamics

Solution

When ice changes into water spontaneously at room temperature, the randomness increases, so entropy increases, which is in accordance with the second law of thermodynamics.
Question 5
1 / -0

Which of the following pairs of a chemical reaction is certain to result in a spontaneous reaction?

A
Exothermic and decreasing disorder

B
Endothermic and increasing disorder

C
Exothermic and increasing disorder

D
Endothermic and decreasing disorder

Solution

A chemical reaction which is exothermic and has an increasing disorder is certain to result in a spontaneous reaction.
Since the reaction is exothermic, $$ \displaystyle \Delta H $$ is negative. Since, the disorder increases, $$ \displaystyle \Delta S $$ is positve.

$$ \displaystyle \Delta G =\Delta H - T\Delta S = -ve - ( +ve \times +ve) =-ve$$

Negative value of $$ \displaystyle \Delta G$$ corresponds to spontaneous process.
Question 6
1 / -0

Match the List-II ans List-II and List III:
List-I List-II List-III
A. $$\Delta \, G>0$$ X. $$\Delta \, S>0$$ 1. Non-spontaneous
B. $$\Delta \, G<0$$ Y. $$\Delta \, S<0$$ 2. Spontaneous
C. $$\Delta \, G=0$$ Z. $$\Delta \, S=0$$ 3. Equilibrium
select the correct answering from the following codes:

A
A-(Y, 1) B-(X, Y) C-(Z, 3)

B
A-(X, 2) B-(Y, 3) C-(Z, 1)

C
A-(X, 3) B-(Y, 1) C-(Z, 2)

D
A-(Y, 1) B-(X, 3) C-(Z, 2)

Solution

According to the third law, the entropy change of a spontaneous process is always greater than zero.
For spontaneous process $$\Longrightarrow \quad \Delta G<0\quad and\quad \Delta S>0$$
For non-spontaneous process $$\Longrightarrow \quad \Delta G>0\quad and\quad \Delta S<0$$
For equilibrium process $$\Longrightarrow \quad \Delta G=0\quad and\quad \Delta S=0$$
Question 7
1 / -0

If $$\triangle H_{vaporisation}$$ of substance $$X(l)$$ (molar mass: $$30\ g/mol)$$ is $$300\ J/g$$ at it's boiling point $$300\ K$$, then molar entropy change for reversible condensation process is____________.

A
$$30\ J/mol.K$$

B
$$-300\ J/mol.K$$

C
$$-30\ J/mol.K$$

D
None of these

Solution

Solution:- (A) $$30 \; {J}/{mol-K}$$
Given:-
$${\Delta{H}}_{vap.} = 300 \; {J}/{g}$$
Molar mass of $$X = 30 \; {g}/{mol}$$
$$\therefore {\Delta{H}}_{molar} = 300 \times 30 = 9000 \; {J}/{mol}$$
Boiling point of substance $$X, \; \left( T \right) = 300 K$$
As we know that,
Molar entropy change, $${\Delta{S}}_{molar} = \cfrac{{\Delta{H}}_{molar}}{T} = \cfrac{9000}{300} = 30 \; {J}/{mol-K}$$
Hence the molar entropy change for the given process will be $$30 \; {J}/{mol-K}$$.
Question 8
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Heat of combustion $$\Delta H^o$$ for $$C(s), H_2(g)$$ and $$CH_4(g)$$ are $$-94,-68$$ and $$-213 kcal/mol$$ respectively. Then $$\Delta H^o$$ for $$C(s) + 2H_2(g) \rightarrow CH_4$$ is:

A
$$-17\, kcal/mol$$

B
$$-111\, kcal/mol$$

C
$$-170\, kcal/mol$$

D
$$-85\, kcal/mol$$

Solution

For reaction,
$$C(s)+2H_2(g)\rightarrow CH_4(g), \Delta H^o=?$$

$$\Delta H^o$$= $$- $$[($$\Delta H^o$$ of combustion of $$CH_4$$) - ($$\Delta H^o$$ of combustion of $$C$$)+ (2 $$\times \Delta H^o $$ of combustion of $$H_2$$)]

$$C + O_2 \rightarrow CO_2; \Delta H^o = - 94\, kcal$$ ... (i)

$$2H_2+O_2 \rightarrow 2H_2O; \Delta H^o = - 68\times 2\, kcal$$ ... (ii)

$$=-[(-213)-(-94+2\times -68)]\, kcal/mol$$

$$=-[-213+230]=-17\,kcal/mol$$
Question 9
1 / -0

$$H_{2(g)} + \dfrac{1}{2} O_{2(g)} \rightarrow H_2O_{(g)}$$ $$\Delta H = 241.8 kJ$$ $$CO_{(g)} + \dfrac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}$$ $$\Delta H = 283 kJ$$
The heat evolved during the combustion of $$112$$ litre of water gas (mixture of equal volume of $${H}_{2}$$ and $$CO$$) is:

A
$$241.8kJ$$

B
$$283kJ$$

C
$$1312kJ$$

D
$$1586kJ$$

Solution

Solution:- (C) $$1312 \; KJ$$
Molecular formula of water gas $$= {H}_{2} + CO$$
Combustion of $${H}_{2}$$-
$${H}_{2} + \cfrac{1}{2} {O}_{2} \longrightarrow {H}_{2}O; \; \Delta{H} = 241.3 \; KJ$$
Combustion of $$CO$$-
$$CO + \cfrac{1}{2} {O}_{2} \longrightarrow C{O}_{2}; \; \Delta{H} = 283 \; KJ$$
$$1$$ mole of $${H}_{2}$$ and $$CO$$ each gives,
$$\Delta{H} = 283 + \left( 241.8 \right) = 524.8 \; KJ$$
$$1$$ mole of $${H}_{2} = 22.4 \; L$$
Similarly,
$$1$$ mole of $$CO = 22.4 \; L$$
$$\therefore$$ Total volume $$= 22.4 + 22.4 = 44.8 \; L$$
Therefore,
Heat evolved during combustion of $$44.8 \; L$$ of water gas $$= 524.8 \; KJ$$
Heat evolved during combustion of $$112 \; L$$ of water gas $$= \cfrac{524.8 \times 112}{44.8} = 1312 \; KJ$$
Hence the heat evolved during the combustion of $$112$$ litre of water gas at STP is $$1312 \; KJ$$.
Question 10
1 / -0

One mole of a real gas is subjected to a process from $$(2\ bar, 30\ lit, 300\ K)$$ to $$(2\ bar, 50\ lit, 400\ K)$$
Given: $$C_{V} = 40\ J/mol\ K$$,
$$C_{P} = 50\ J/mol/ K$$
Calculate $$\Delta U$$.

A
$$4000\ J$$

B
$$2000\ J$$

C
$$1000\ J$$

D
$$5000\ J$$

Solution

Change in internal energy is written as:
$$\Delta U = nC_v \Delta T$$
$$\Delta U = 1\times 40\times (400-300)$$
$$\Delta U = 4000J$$

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Re-Attempt Weekly Quiz Competition

List-I	List-II	List-III
A. $$\Delta \, G>0$$	X. $$\Delta \, S>0$$	1. Non-spontaneous
B. $$\Delta \, G<0$$	Y. $$\Delta \, S<0$$	2. Spontaneous
C. $$\Delta \, G=0$$	Z. $$\Delta \, S=0$$	3. Equilibrium

Thermodynamics Test - 32

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Thermodynamics Test - 32

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SHARING IS CARING

Question 1

A process which is exothermic and evolves a lot of heat

A process which is slow and reversible

A process which takes place only in presence of a catalyst

A process that occurs without any input from the surroundings

Question 2

$$0kJ$$

$$30kJ$$

$$10kJ$$

$$20kJ$$

Solution

Question 3

$$-5.7$$

$$5.7$$

$$11.4$$

$$-11.4$$

Solution

Question 4

Second law of thermodynamics

First law of thermodynamics

Third law of thermodynamics

Zeroth law of thermodynamics

Solution

Question 5

Exothermic and decreasing disorder

Endothermic and increasing disorder

Exothermic and increasing disorder

Endothermic and decreasing disorder

Solution

Question 6

A-(Y, 1) B-(X, Y) C-(Z, 3)

A-(X, 2) B-(Y, 3) C-(Z, 1)

A-(X, 3) B-(Y, 1) C-(Z, 2)

A-(Y, 1) B-(X, 3) C-(Z, 2)

Solution

Question 7

$$30\ J/mol.K$$

$$-300\ J/mol.K$$

$$-30\ J/mol.K$$

None of these

Solution

Question 8

$$-17\, kcal/mol$$

$$-111\, kcal/mol$$

$$-170\, kcal/mol$$

$$-85\, kcal/mol$$

Solution

Question 9

$$241.8kJ$$

$$283kJ$$

$$1312kJ$$

$$1586kJ$$

Solution

Question 10

$$4000\ J$$

$$2000\ J$$

$$1000\ J$$

$$5000\ J$$

Solution

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