Thermodynamics Test

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Thermodynamics Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

For the reaction of one mole zinc dust with one mole sulphuric acid in a bomb calorimeter, $$\triangle U$$ and w correspond to?

A
$$\triangle U<0,w=0$$

B
$$\triangle U<0,w<0$$

C
$$\triangle U>0,w=0$$

D
$$\triangle U>0,w>0$$

Solution

$$Zn(s)+H_{2}SO_{4}(g)\rightleftharpoons ZnSO_{4}(s)+H_{2}(g)$$

$$\Delta _{n_{gas}}=0$$

$$\therefore w=\Delta _{n_{gas}}RT=0$$

$$\Delta U=w+q$$

Bomb calorimeter is commonly used to find the heat of combustion of organic substances which consists of a sealed combustion chamber, called a bomb. If a process is run in a sealed container then no expansion or compression is allowed, so w = 0 and ∆U = q.

$$\therefore q<0$$
$$\therefore \Delta U<0$$

Hence, the correct option is $$\text{A}$$
Question 2
1 / -0

For $$ { Zn }^{ 2+ }/Zn $$ ,
$$ { E }^{ \circ }=-0.76\quad V $$,
for $$ { Ag }^{ + }/Ag\quad { E }^{ \circ }=\quad 0.799\quad V $$.
The correct statement is:

A

Zn undergoes reduction and Ag is oxidized

B
in the reaction Zn getting reduced, Ag getting oxidized is spontaneous

C
Zn undergoes oxidation $$Ag^+$$ gets reduced

D
no suitable answer

Solution

For feasible reaction,
$$E_{cell} > 0$$
$$E^o Ag^+ / Ag > E^o \, zn^{2+}/ zn$$
$$\therefore Ag^+$$ gets reduced and $$zn^-$$ get oxidised to $$zn^{2+}$$
Question 3
1 / -0

Suppose that a reaction has $$\Delta H=40\ kJ$$ and $$\Delta S=50\ kJ/ K$$. At what temperature range will it change from spontaneous to non-spontaneous?

A
$$0.8\ K$$ to $$1\ K$$

B
$$799\ K$$ to $$800\ K$$

C
$$800\ K$$ to $$801\ K$$

D
$$799\ K$$ to $$801\ K$$

Solution

$$\Delta G=\Delta H-T\Delta S$$
$$\Delta H=$$40 kJ
$$\Delta S=$$50 kJ/K
if $$\Delta G$$<0, process is spontaneous
if $$\Delta G$$>0, process is non spontaneous
if $$\Delta G$$=0, process is in equilibrium
For T$$=0.8$$K
$$\Delta G=40-{0.8}\times{50}=0$$
$$\Delta G$$=0
The process is in equilibrium
For T$$=1 $$K
$$\Delta G=40-{1}\times{50}=-10$$
The process is in spontaneous
Question 4
1 / -0

Two liquids are at $$40^oC$$ and $$30^oC.$$ When they are mixed in equal masses , the temperature of the mixture is $$36^oC.$$ Ratio of their specific heats is

A
$$3:2$$

B
$$2:3$$

C
$$4:3$$

D
$$3:4$$

Solution

Let liquid - 1 has specific heat $$s_{1}$$ at temp.$$40^{\circ}C$$

Liquid - 2 has specific heat $$s_{2}$$ at temp. $$30^{\circ}C$$

common temp. is $$36^{\circ}C$$ and both have equal mass (suppose $$m$$)

heat lost by liquid-1 is

$$Q_{1}=ms_{1}\Delta t=ms_{1}(40-36)=4ms_{1}$$

heat gained by liquid - 2 is $$Q_{2}= ms_{2}\Delta t^{'}$$

$$\Rightarrow Q_{2}=ms_{2}\times (36-30)= 6ms_{2}$$

both should be same according to the conservation of energy

$$Q_{1}=Q_{2}$$

or $$4ms_{1}=6ms_{2}$$

$$\Rightarrow \dfrac{S_{1}}{S_{2}}=\dfrac{6}{4}=\dfrac{3}{2}$$

$$S_{1} S_{2}=3:2$$
Question 5
1 / -0

The value closest to the thermal velocity of a Helium atom at room temperature (300 K) in $$ms^{-1}$$ is : [$$k_B = 1.4 \times 10^{-23} \ J/K ; \ m_{He} = 7\times 10^{-27} \ kg$$ ]

A
$$1.3 \times 10^4$$

B
$$1.3 \times 10^3$$

C
$$1.3 \times 10^5$$

D
$$1.3 \times 10^2$$

Solution

Given,
$$T=300K$$
$$m=7\times 10^{-27}kg$$
$$k_B=1.4\times 10^{-23}J/K$$
The thermal velocity of helium is given by
$$v=\sqrt{\dfrac{3k_BT}{m}}$$
$$v=\sqrt{\dfrac{3\times 1.4\times 10^{-23}\times 300}{7\times 10^{-27}}}$$
$$v=1.3\times 10^3m/s$$
The correct option is B.
Question 6
1 / -0

For the reaction; $${ FeCO }_{ 3(s) }\longrightarrow { Fe }O_{ (s) }+C{ O }_{ 2 }(g)$$, $$\Delta H=82.8\ kJ$$ at $${ 25 }^{ 0 }C$$. What is $$\Delta U$$ at $${ 25 }^{ 0 }C$$?

A
$$82.8 kJ$$

B
$$80.32 kJ$$

C
$$-2394.77 kJ$$

D
$$85.28 kJ$$

Solution

Solution:- (B) $$80.32 \; kJ$$

$${FeC{O}_{3}}_{\left( s \right)} \longrightarrow {FeO}_{\left( s \right)} + {C{O}_{2}}_{\left( g \right)}$$

From first law of thermodynamics-

$$\Delta{U} = \Delta{H} - \Delta{n} RT$$

$$\Delta{H} = 82.8 \; KJ$$

$$T = 25 ℃ = \left( 25 + 273 \right) K = 298 K$$

$$R = 8.314 \; J \; {mol}^{-1} {K}^{-1} =8.314 \times {10}^{-3} \; kJ \; {mol}^{-1} {K}^{-1}$$

$$\Delta{n} = {n}_{p} - {n}_{r} = \left( 1 \right) - 0 = 1$$

$$\therefore \Delta{U} = 82.8 - \left( 1 \times 8.314 \times {10}^{-3} \times 298 \right)$$

$$\Rightarrow \Delta{U} = 82.8 - 2.48 = 80.32 \; kJ$$

Hence for th given reaction, $$\Delta{U}$$ at $$25 ℃$$ is $$80.32 \; kJ$$.
Question 7
1 / -0

For a reaction to occur spontaneously:

A
$$\Delta S$$ must be negative

B
$$(-\Delta H+T \Delta S)$$ must be positive

C
$$\Delta H+T\Delta S$$ must be negative

D
$$\Delta H$$ must be negative

Solution

A spontaneous reaction is a reaction that occurs in a given set of conditions without intervention.Spontaneous reaction are accompanied by an increase in overall entropy, or disorder. If the Gibbs Free Energy is negative, then the reaction is spontaneous, and if it is positive, then it is nonspontaneous.
conclusion: hence the option (A) is correct.
Question 8
1 / -0

$$\Delta S $$ for $$4Fe_(s)+3O_{2(g)}\rightarrow 2Fe_2O_{3(s)}$$ is $$550J/mol/K$$. The process is found to be spontaneous even at $$298K$$ $$[\Delta H=-1650kJ]$$

A
$$\Delta S_{total}=-2000J$$

B
$$\Delta S_{total}=+1650J$$

C
$$\Delta S_{total }=+4980J$$

D
$$\Delta S_total =-4980J$$

Solution
Question 9
1 / -0

The incorrect criterion for the spontaneity of a process is:

A
$$\Delta S_{system }<0$$

B
$$\Delta S_{system }+\Delta S_{surrounding }>0$$

C
$$\Delta S_{system }>\Delta S_{surrounding }$$

D
$$\Delta H_{system }<0$$ and $$\Delta S_{system}>0$$

Solution

For spontaneity of a process
$${ \Delta S }_{ system }+{ \Delta S }_{ sourroundings }>0$$
$${ \Delta S }_{ system }>0$$
$${ \Delta S }_{ surroundings }<0$$

Ans :- Option A.
Question 10
1 / -0

$$0.1 \,{m^3}$$ of water at $${80^ \circ }$$ is mixed with $$0.3 {m^3}$$ of water at $${60^ \circ }$$. The final temperature of mixture is :

A
$${80^ \circ }C$$

B
$${70^ \circ }C$$

C
$${60^ \circ }C$$

D
$${75^ \circ }C$$

Solution

$${ m }_{ 1 }{ s }_{ 1 }{ dt }_{ 1 }={ m }_{ 2 }{ s }_{ 2 }{ dt }_{ 2 }$$
$$P\left[ 0.1 \right] \times 1\times \left[ 80-t \right] =P\left[ 0.3 \right] \times \left[ 1 \right] \times \left[ t-60 \right] $$
$$80-t=3t-180$$
$$4t=260=t=6{ 5 }^{ 0 }c$$

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Thermodynamics Test - 38

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Thermodynamics Test - 38

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Question 1

$$\triangle U<0,w=0$$

$$\triangle U<0,w<0$$

$$\triangle U>0,w=0$$

$$\triangle U>0,w>0$$

Solution

Question 2

Zn undergoes reduction and Ag is oxidized

in the reaction Zn getting reduced, Ag getting oxidized is spontaneous

Zn undergoes oxidation $$Ag^+$$ gets reduced

no suitable answer

Solution

Question 3

$$0.8\ K$$ to $$1\ K$$

$$799\ K$$ to $$800\ K$$

$$800\ K$$ to $$801\ K$$

$$799\ K$$ to $$801\ K$$

Solution

Question 4

$$3:2$$

$$2:3$$

$$4:3$$

$$3:4$$

Solution

Question 5

$$1.3 \times 10^4$$

$$1.3 \times 10^3$$

$$1.3 \times 10^5$$

$$1.3 \times 10^2$$

Solution

Question 6

$$82.8 kJ$$

$$80.32 kJ$$

$$-2394.77 kJ$$

$$85.28 kJ$$

Solution

Question 7

$$\Delta S$$ must be negative

$$(-\Delta H+T \Delta S)$$ must be positive

$$\Delta H+T\Delta S$$ must be negative

$$\Delta H$$ must be negative

Solution

Question 8

$$\Delta S_{total}=-2000J$$

$$\Delta S_{total}=+1650J$$

$$\Delta S_{total }=+4980J$$

$$\Delta S_total =-4980J$$

Solution

Question 9

$$\Delta S_{system }<0$$

$$\Delta S_{system }+\Delta S_{surrounding }>0$$

$$\Delta S_{system }>\Delta S_{surrounding }$$

$$\Delta H_{system }<0$$ and $$\Delta S_{system}>0$$

Solution

Question 10

$${80^ \circ }C$$

$${70^ \circ }C$$

$${60^ \circ }C$$

$${75^ \circ }C$$

Solution

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