Thermodynamics Test

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Thermodynamics Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

To which portion of the heating curve for water does $$C=1.00 \ cal/{g}^{o}C$$ apply?

A
$$A-B$$

B
$$B-E$$

C
$$C-D$$

D
$$A-F$$
Solution
- After all ice is melted to liquid at 'C' the temperature of liquid increases as heat is observed.
- It is then possible to calculate heat absorbed by using specific heat of water in the region CD,
Question 2
1 / -0

For a thermodynamics process to be reversible, the temperature difference between hot body and the working substance should be

A
zero

B
minimunm

C
maximum

D
infinity

Solution

In thermodynamics, a reversible process is a process whose direction can be "reversed" by inducing infinitesimal changes to some property of the system via its surroundings, while not increasing entropy. Throughout the entire reversible process, the system is in thermodynamic equilibrium with its surroundings, which means that the hot body and surroundings are at same temperature.
Question 3
1 / -0

In a thermodynamic process, a system absorbs 2 k cal of heat and at the same time does 800 J of work. The change in internal energy of the system is

A
7600 J

B
2000 J

C
4800 J

D
-7600 J

Solution

In a thermodynamic process, a system absorbs 2 k cal of heat and at the same time does 800 J of work
By calorie to joules conversion
we have $$1 cal =4.2J$$
$$\therefore$$$$2Kcal=8400 J$$
Hence by first law of thermodynamics
we have,$$\delta U=\delta H -W=8400-800=7600J$$
Question 4
1 / -0

The criterion for a spontaneous process is :

A
$$\triangle G > 0$$

B
$$\triangle G < 0$$

C
$$\triangle G = 0$$

D
$$\triangle S_{total} < 0$$

Solution

For a spontaneous process to occur, the free energy should always decrease from its initial value.

So, $$\Delta G<0$$.
Question 5
1 / -0

Consider the following changes
$$M\left( s \right) \longrightarrow M\left( g \right) $$ ...(1)
$$M\left( g \right) \longrightarrow { M }^{ 2+ }\left( g \right) +2{ e }^{ - }$$ ....(2)
$$M\left( g \right) \longrightarrow { M }^{ + }\left( g \right) +{ e }^{ - }$$ ....(3)
$${ M }^{ + }\left( g \right) \longrightarrow { M }^{ 2+ }\left( g \right) +{ e }^{ - }$$ ...(4)
$$M\left( g \right) \longrightarrow { M }^{ 2+ }\left( g \right) +2{ e }^{ - }$$ ...(5)
The second ionisation energy of $$M$$ could be determined from the energy values associated with:

A
$$1+2+4$$

B
$$1+5-3$$

C
$$2+3-4$$

D
$$5-3$$

Solution

The amount of energy required to take out an electron from the monopositive cation is called second ionisation energy
$$M\left( g \right) \longrightarrow { M }^{ 2+ }\left( g \right) +2{ e }^{ - }$$ (v)
$$M\left( g \right) \longrightarrow { M }^{ + }\left( g \right) +{ e }^{ - }$$ (iii)
On subtracting equation (iii) from equation (v) we get,
$$M\longrightarrow { M }^{ 2+ }+{ e }^{ - }$$.
Question 6
1 / -0

In which of the processes, does the internal energy of the system remain constant?

A
Adiabatic

B
Isochoric

C
Isobaric

D
Isothermal

Solution

For an ideal gas, in an isothermal process the change in internal energy is 0 as internal energy is function of temperate,but temperature remains constant.
Question 7
1 / -0

Thermodynamic condition for irreversible spontaneous process at constant '$$T$$' and '$$P$$' is:

A
$$\Delta G< 0$$

B
$$\Delta G=0$$

C
$$\Delta G> 0$$

D
Both (B) and (C)

Solution

For spontaneous process $$\Delta G<0$$ and for non spontaneous process $$\Delta G>0$$.

Option A is correct.
Question 8
1 / -0

For the reaction $$N_2+3X_2 \rightarrow 2NX_3$$ where $$X = F, Cl$$ (the average bond energies are $$F-F = 155 \ kJ \ mol^{-1}$$, $$N-F = 272 \ kJ \ mol^{-1}$$, $$Cl-Cl=242\ kJ \ mol^{-1}$$ , $$N-Cl=200\ kJ \ mol^{-1}$$ and $$N \equiv N = 941 kJ\, mol^{-1}$$). The heats of formation of $$NF_3$$ and $$NCl_3$$ in $$kJ\ mol^{-1}$$, respectively, are closest to:

A
$$- 226$$ and $$+ 467$$

B
$$+ 226$$ and $$-467$$

C
$$-151$$ and $$+ 311$$

D
$$+ 151$$ and $$-311$$

Solution

$$N_2 + 3X_2 \rightarrow 2NX_3$$

Energy to break 1 mole of $$N_2$$ bond $$= 941 \ kJ mol^{-1}$$
Energy to break 3 mole of $$ F_2$$ bond $$= 3 \times 155 \ kJ \ mol^{-1}$$
Energy released caused of bond formation of 2 moles of 3 bonds $$N-F = 2 \times 3 \times 272$$

Heat of formation of $$NF_3$$ by Hess law $$= 941+3 \times 155 - (2 \times 3 \times 272) = -226 \ kJ \ mol^{-1}$$

Similarly for $$NCl_3$$,

Just replacing the necessary values in previous calculations, heat of formation of $$NCl_3 = 941 + 3 \times 242 - (2 \times 3 \times 200) = +467 \ kJ mol^{-1}$$.

Hence, option $$A$$ is correct.
Question 9
1 / -0

The molar enthalpy change for $$H_2O(1)\rightleftharpoons H2O(g)$$ at $$373$$K and $$1$$ atm is $$41$$kJ/mol. Assuming ideal behavior, the internal energy change for vaporization of $$1$$ mol of water at $$373$$K and $$1$$ atm in kJ $$mol^{-1}$$ is $$?$$

A
$$30.2$$

B
$$41.0$$

C
$$48.1$$

D
$$37.9$$

Solution

$$W=-nRT=1(8.314\times 373)\times 10^{-3}=-3.10 KJ$$
$$q=41KJ$$
$$\Delta U=37.9KJ$$
Question 10
1 / -0

Under which of the following conditions is the relation,
$$\Delta H=\Delta U+P\Delta V$$ valid for closed system?

A
Constant pressure

B
Constant temperature

C
Constant temperature and pressure

D
Constant temperature, pressure and composition

Solution

$$H=C+PV\\ \triangle H=\triangle C+P\triangle V+V\triangle P$$
If $$\triangle P=0$$, i.e. pressure is constant
$$\triangle H=\triangle C+P\triangle V$$
Hence, the above relation is valid at constant pressure and temperature.

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Thermodynamics Test - 26

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Thermodynamics Test - 26

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Question 1

$$A-B$$

$$B-E$$

$$C-D$$

$$A-F$$

Solution

Question 2

zero

minimunm

maximum

infinity

Solution

Question 3

7600 J

2000 J

4800 J

-7600 J

Solution

Question 4

$$\triangle G > 0$$

$$\triangle G < 0$$

$$\triangle G = 0$$

$$\triangle S_{total} < 0$$

Solution

Question 5

$$1+2+4$$

$$1+5-3$$

$$2+3-4$$

$$5-3$$

Solution

Question 6

Adiabatic

Isochoric

Isobaric

Isothermal

Solution

Question 7

$$\Delta G< 0$$

$$\Delta G=0$$

$$\Delta G> 0$$

Both (B) and (C)

Solution

Question 8

$$- 226$$ and $$+ 467$$

$$+ 226$$ and $$-467$$

$$-151$$ and $$+ 311$$

$$+ 151$$ and $$-311$$

Solution

Question 9

$$30.2$$

$$41.0$$

$$48.1$$

$$37.9$$

Solution

Question 10

Constant pressure

Constant temperature

Constant temperature and pressure

Constant temperature, pressure and composition

Solution

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