Thermodynamics Test

Result

Thermodynamics Test - 17

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Actual flame temperature is always lower than the adiabatic flame temperature, because there is ______________.

A
no possibility of obtaining complete combustion at high temperature.

B
always loss of heat from the flame.

C
both (a) and (b).

D
neither (a) nor (b).

Solution

The constant volume adiabatic flame temperature is the temperature that results from a complete combustion process that occurs without any work, heat transfer or changes in kinetic or potential energy. The constant pressure adiabatic flame temperature is the temperature that results from a complete combustion process that occurs without any heat transfer or changes in kinetic or potential energy. Its temperature is lower than the constant volume process because some of the energy is utilized to change the volume of the system (i.e., generate work).
Question 2
1 / -0

The internal energy of a compressed real gas, as compared to that of the normal gas at the same temperature, is

A
less

B
more

C
sometimes less, sometimes more

D
none of these

Solution

The internal energy wil be less due to the short range repulsive forces present between the molecules of the real gas.
Question 3
1 / -0

Two bodies at different temperatures are mixed in a calorimeter. Which of the following quantities remains conserved?

A
sum of the temperatures of the two bodies

B
total heat of the two bodies

C
total internal energy of the two bodies

D
internal energy of each body

Solution

There is no heat lost since the calorimeter is insulated .
If we consider both the liquids together as one system , then the work done by the system is zero.

From first law of thermodynamics,
$$\Delta U = Q - W$$
$$\Delta U = 0$$

Since there is no change in internal energy, the internal energy of the system remains constant.
Question 4
1 / -0

Hess's law is applicable for the determination of heat of:

A
reaction

B
transition

C
formation

D
all of the above

Solution

Hess's law of constant heat summation deals with the enthalpy change for a chemical reaction. The chemical reaction can involve any chemical change. It may include formation or transition.
Question 5
1 / -0

Which specific process has negative value of specific heat?

A
Saturated vapours

B
Ice

C
Water

D
Vapours

Solution

In case of saturated vapors, when a certain quantity of heat is removed, the temperature increases. Hence, the specific heat is negative for saturated vapors.
Question 6
1 / -0

The Gibbs free energy change for the reaction for which $$\displaystyle \Delta H=-393.4kJ$$ and $$ \Delta S=-2.9{ JK }^{ -1 }{ mol }^{ -1 }$$ at 298K is:

A
$$\displaystyle -352.5kJ$$

B
$$\displaystyle -392.5kJ$$

C
$$\displaystyle +352.5kJ$$

D
$$\displaystyle+392.5kJ$$

Solution

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

Given, $$\Delta H=-393.4\ kJ/mol$$ and $$\Delta S=-2.9\ JK^{-1}mol^{-1}$$

$$\Delta G = -393400-298\times 2.9=-392535.8\ J\approx -392\ kJ$$

Hence, option B is correct.
Question 7
1 / -0

The internal energy of a gram-molecule of an ideal gas depends on

A
pressure alone

B
volume alone

C
temperature alone

D
both pressure as well as temperature

Solution

The internal energy of an ideal gas is given by the expression : $$U = C_{v}T$$ .
From the above expression its clear that internal energy depend only on temperature.
Question 8
1 / -0

Two moles of an ideal gas are compressed isothermally $$\displaystyle \left( { 100 }^{ o }C \right) $$ and reversibly from a pressure of 10 atm to 25 atm. The value of Gibb's free energy change is:

A
-3.684 kJ

B
+5.684 kJ

C
+3.684 kJ

D
-5.684 kJ

Solution

For isothermal reversible process:
$$\displaystyle { W }_{ rev }=-2.303nRT\log { \left( \dfrac{ { P }_{ 1 } }{ { P }_{ 2 } } \right) } $$
$$\displaystyle =-2.303\times 2\times 8.314\times 373\times \log { \left(\dfrac { 10 }{ 25 } \right) } $$$$\displaystyle =+5684.1 J$$
$$\displaystyle { W }_{ rev }=+5.684kJ$$
Since, $$\displaystyle { W }_{ rev }$$ is a measure of free energy,
i.e., $$\displaystyle \Delta G={ W }_{ rev }={ W }_{ max }$$
$$\displaystyle \therefore \Delta G=+5.684kJ$$
Question 9
1 / -0

For a given reaction, $$ \Delta H=35.5\ kJ\ mol^{-1}$$ and $$ \Delta S=83.6\ J\ K^{-1}\ mol^{-1}. The reaction is spontaneous at:
(Assume that $$ \Delta H$$ and $$ \Delta S$$ so not vary with temperature)

A

$$T<425K$$

B
$$T>425K$$

C
all temperatures

D
$$T>298K$$
Question 10
1 / -0

Determine enthalpy change for, $$C_3H_8 (g )+ H_2(g)\longrightarrow C_2H_6(g)+ CH_4 (g)$$ at $$25^{\circ}C$$. using heat of combustion values under standard conditions.
Compounds
$$H_2(g)$$
$$CH_4(g)$$
$$C_2H_6(g)$$
$$C_{graphic}$$
$$\Delta H^{\circ}$$ in kJ/mol
-285.8
-890.0
-1560.0
-393.5
The standard heat of formation of $$C_3H_8 (g )$$ is $$-103.8 \: kJ \: mol^{-1}$$.

A
$$ -55.7 \,kJ$$

B
$$ -75.5 \,kJ$$

C
$$ -55.7 \,J$$

D
None of these

Solution

Given.
           $$H_2(g)+\frac{1}{2}O_2(g)\longrightarrow H_2O(g); \:                     \Delta H^{\circ}= -285.8 \,kJ$$           ...(i)
           $$CH_4(g)+ 2O_2(g) \longrightarrow CO_2(g)+ 2H_2O(g) \:   \Delta H^{\circ}= -890.0 \,kJ$$           ...(ii)
$$C_2H_6(g)+ (7/2)O_2(g)\longrightarrow 2CO_2(g)+ 3H_2O(g); \:\Delta H^{\circ}= -1560.0 \,kJ$$         ...(iii)
          $$3C(s)+4H_2(g)\longrightarrow C_3H_8(g);\:                      \Delta H^{\circ}= -103.8\,kJ$$           ...(iv)
                 $$C(s)+ O_2(g)\longrightarrow CO_2(g); \:                     \Delta H^{\circ}= -393.5 \,kJ$$          ... (v)
By inspection method
$$5 \times (i) + 3 \times (v) - [(ii) + (iii) + (iv)]$$
$$H_2(g)+ C_3H_8(g) \longrightarrow CH_4(g)+ C_2H_8(g); \:                              \Delta H^{\circ}= -55.7 \,kJ$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Compounds	$$H_2(g)$$	$$CH_4(g)$$	$$C_2H_6(g)$$	$$C_{graphic}$$
$$\Delta H^{\circ}$$ in kJ/mol	-285.8	-890.0	-1560.0	-393.5

Thermodynamics Test - 17

Result

Thermodynamics Test - 17

Score

-

Rank

-

SHARING IS CARING

Question 1

no possibility of obtaining complete combustion at high temperature.

always loss of heat from the flame.

both (a) and (b).

neither (a) nor (b).

Solution

Question 2

less

more

sometimes less, sometimes more

none of these

Solution

Question 3

sum of the temperatures of the two bodies

total heat of the two bodies

total internal energy of the two bodies

internal energy of each body

Solution

Question 4

reaction

transition

formation

all of the above

Solution

Question 5

Saturated vapours

Ice

Water

Vapours

Solution

Question 6

$$\displaystyle -352.5kJ$$

$$\displaystyle -392.5kJ$$

$$\displaystyle +352.5kJ$$

$$\displaystyle+392.5kJ$$

Solution

Question 7

pressure alone

volume alone

temperature alone

both pressure as well as temperature

Solution

Question 8

-3.684 kJ

+5.684 kJ

+3.684 kJ

-5.684 kJ

Solution

Question 9

$$T<425K$$

$$T>425K$$

all temperatures

$$T>298K$$

Question 10

$$ -55.7 \,kJ$$

$$ -75.5 \,kJ$$

$$ -55.7 \,J$$

None of these

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.