Thermodynamics Test

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Thermodynamics Test - 57

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Weekly Quiz Competition

Question 1
1 / -0

How much energy is needed to convert $$100\ g$$ of ice at $$263\ K$$ to liquid water at a temperature of $$283\ K$$?
$${C}_{ice}=0.49\ cal/({g}^{o}C)$$
$${C}_{water}=1.00\ cal/({g}^{o}C)$$
$$\Delta {H}_{fus}= 79.8\ cal/{g}$$
$$\Delta {H}_{vap}=540\ cal/g$$

A
$$9470\ cal$$

B
$$3288\ cal$$

C
$$2288\ cal$$

D
$$727.3\ cal$$

Solution

ice $$ \rightarrow $$ liquid water
$${C}_{ice}=0.49\ cal/({g}^{o}C)$$
$${C}_{water}=1.00\ cal/({g}^{o}C)$$
$$\Delta {H}_{fus}= 79.8\ cal/{g}$$
$$\Delta {H}_{vap}=540\ cal/g$$
ice $$ \rightarrow $$ melt (fusicm) $$ \rightarrow $$ water
$$(263\,k)$$ At $$0^{\circ}c(237\,k)$$ $$ 283\,k$$
ENERGY $$\displaystyle = \int_{263}^{273}C_{ice}dt+\Delta H_{fus}+\int_{273}^{283}C_{water}dt $$
$$ = 0.49(273-263)+79.8+1(283-273) $$
$$ = 94.7\,cal\,g $$
For 100 g ice energy
$$ = 94.7\times 100\,cal $$
$$ = 9470\,cal $$
Question 2
1 / -0

For a reaction to be spontaneous, the sign on delta G should be :

A
positive

B
there should be no sign

C
negative

D
spontaneity is not related to Gibbs Free Energy

E
positive or Negative

Solution

$$ \triangle G = negative $$ for spontaneous reaction.
$$ \triangle G = Positive $$ for non-spontaneous reaction.
$$ \triangle G = 0 $$ for equilibrium.
Question 3
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Which of the following is most likely to produce a spontaneous reaction?

A
Negative Enthalpy

B
Positive Enthalpy

C
Negative Entropy

D
Positive Entropy

E
Negative Enthalpy and positive Entropy

Solution

For the spontaneous reaction, the negative enthalpy and positive entropy is most likely.
As for spontaneous reaction the value of $$\Delta G $$ must be negative.
Hence, as per formula,
$$\Delta G= \Delta H - T \Delta S$$,
So, if $$\Delta H $$ and $$\Delta S$$ are negative and positive respectively then the value of the $$\Delta G $$ must be negative whatever the temperature.
Question 4
1 / -0

If the gibbs free energy is negative than reaction will be?

A
always positive

B
sometimes negative

C
non-spontaneous

D
spontaneous

E
None of the above

Solution

(D) : Always negative.
$$\Delta G = 0$$, it means the reaction is equilibrium at standard conditions.
A negative value of $$\Delta G $$, means spontaneous.
A positive value of $$\Delta G $$, means non-spontaneous.
Question 5
1 / -0

An ideal mono-atomic gas of given mass is heated at constant pressure. In this process, the fraction of supplied energy used for the increase of the internal energy of the gas is

A
$$\dfrac{3}{8}$$

B
$$\dfrac{3}{5}$$

C
$$\dfrac{3}{4}$$

D
$$\dfrac{2}{5}$$

Solution

Let the change in temperature of the gas be $$\Delta T$$.
Thus, the amount of heat supplied under constant pressure is $$Q = C_P\Delta T$$
The change in internal energy is $$\Delta U = C_V \Delta T$$
Thus, the fraction of energy used for internal energy is $$\dfrac{\Delta U}{Q}= \displaystyle \frac{C_V}{C_P}$$
For a mono-atomic ideal gas, $$C_V = 3R/2$$ and $$C_P = 5R/2$$
Thus, $$\displaystyle \frac{\Delta U}{Q} = \frac{C_V}{C_P} = \frac{3R/2}{5R/2}=\frac{3}{5}$$
Question 6
1 / -0

$$Mg(s) \rightarrow \rightarrow \rightarrow Mg^{2+}$$ ; Energy = A KJ/mol.
$$Cl_{2}(g) \rightarrow \rightarrow \rightarrow 2Cl^{-}$$ ; Energy = B KJ/mol.
What is the $$\triangle H_{f}$$ of $$MgCl_{2}$$ if lattice enthalpy involved is C KJ/mol ?

A
$$A+B+C$$ KJ/mol.

B
$$A-B+C$$ KJ/mol.

C
$$A+B-C$$ KJ/mol.

D
$$-(A+B+C)$$ KJ/mol.

Solution

$$ Mg(s)\rightarrow Mg(g)\rightarrow Mg^{ 2+ }(g) $$ Energy = AkJ/mol
$$ Cl_{ 2 }\rightarrow 2Cl(g)\rightarrow 2Cl^{ 2- }(g) $$ Energy = BkJ/mol
Adding two equations:
$$ Mg+{ Cl }_{ 2 }\rightarrow Mg{ Cl }_{ 2 } $$ $$ U $$=CkJ/mol
$$ \Delta { H }_{ f }=\Delta { H }_{ sub }+IE+0.5\Delta { H }_{ diss }+{ E }_{ ea }+{ U }_{ lattice\quad enthalpy } $$
$$ \Delta { H }_{ f }$$ = A+B+C KJ/mol
Question 7
1 / -0

For $$A\rightarrow B,\, \Delta H=4\,\text{kcal mol}^{-1}$$, $$\Delta S=10 \,\text{cal mol}^{-1}K^{-1}$$. Reaction is spontaneous when temperature is:

A
$$400$$K

B
$$300$$K

C
$$500$$K

D
None of these

Solution

For a spontaneous reaction $$\Delta H-T\Delta S<0$$

Reaction will be spontaneous if $$T>\dfrac{\Delta H}{\Delta S}$$

Reaction will be spontaneous if $$T>\dfrac{4000}{10}$$

$$T>400\ K$$
Question 8
1 / -0

For spontaneity of cell, which is correct?

A
$$\Delta G=0, \ \Delta H=0$$

B
$$\Delta G=-ve, \ \Delta H=0$$

C
$$\Delta G=+ve,\ \Delta H=0$$

D
$$\Delta G=-ve$$

Solution

For any spontaneous process, Gibbs free energy should be negative. $$\Delta G <0$$ and $$E_{cell}$$ should be positive for this case.
Question 9
1 / -0

The process is spontaneous at the given temperature, if:

A
$$\Delta H$$ is $$+ve$$ and $$\Delta S$$ is $$-ve$$

B
$$\Delta H$$ is $$-ve$$ and $$\Delta S$$ is $$+ve$$

C
$$\Delta H$$ is $$+ve$$ and $$\Delta S$$ is $$+ve$$

D
$$\Delta H$$ is $$+ve$$ and $$\Delta S$$ is equal to zero

Solution

Any process will be spontaneous if $$\Delta S=+ve$$ and $$\Delta H=-ve$$ and this condition make $$\Delta G=-ve$$. If $$\Delta G=\Delta H+T\Delta S$$ is negative then the reaction is spontaneous.

Hence, option B is correct.
Question 10
1 / -0

Consider the below diagram of heat transfer and work transfer for a system. What will be the first law equation for the below system?

A
$$(Q1-Q2) = E+ ( W2 + W3 + W1 )$$

B
$$(Q1 + Q2) = E + ( W2 + W3 + W1 )$$

C
$$(Q1 - Q2) = E + ( W2 + W3 - W1 )$$

D
None of the above

Solution

First law of thermodynamics is:
$$Internal\ energy,\ E=q+w$$
work done by the system and heat released by the system is negative
Therefore,
$$E=(Q_1-Q_2)+(w_1-w_2-w_3)$$
or $$(Q_1-Q_2)=E+(-w_1+w_2+w_3)$$
Therefore option C is correct

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Thermodynamics Test - 57

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Thermodynamics Test - 57

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Question 1

$$9470\ cal$$

$$3288\ cal$$

$$2288\ cal$$

$$727.3\ cal$$

Solution

Question 2

positive

there should be no sign

negative

spontaneity is not related to Gibbs Free Energy

positive or Negative

Solution

Question 3

Negative Enthalpy

Positive Enthalpy

Negative Entropy

Positive Entropy

Negative Enthalpy and positive Entropy

Solution

Question 4

always positive

sometimes negative

non-spontaneous

spontaneous

None of the above

Solution

Question 5

$$\dfrac{3}{8}$$

$$\dfrac{3}{5}$$

$$\dfrac{3}{4}$$

$$\dfrac{2}{5}$$

Solution

Question 6

$$A+B+C$$ KJ/mol.

$$A-B+C$$ KJ/mol.

$$A+B-C$$ KJ/mol.

$$-(A+B+C)$$ KJ/mol.

Solution

Question 7

$$400$$K

$$300$$K

$$500$$K

None of these

Solution

Question 8

$$\Delta G=0, \ \Delta H=0$$

$$\Delta G=-ve, \ \Delta H=0$$

$$\Delta G=+ve,\ \Delta H=0$$

$$\Delta G=-ve$$

Solution

Question 9

$$\Delta H$$ is $$+ve$$ and $$\Delta S$$ is $$-ve$$

$$\Delta H$$ is $$-ve$$ and $$\Delta S$$ is $$+ve$$

$$\Delta H$$ is $$+ve$$ and $$\Delta S$$ is $$+ve$$

$$\Delta H$$ is $$+ve$$ and $$\Delta S$$ is equal to zero

Solution

Question 10

$$(Q1-Q2) = E+ ( W2 + W3 + W1 )$$

$$(Q1 + Q2) = E + ( W2 + W3 + W1 )$$

$$(Q1 - Q2) = E + ( W2 + W3 - W1 )$$

None of the above

Solution

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