Thermodynamics Test

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Thermodynamics Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

The volume of a gas is reduced adiabatically to $$\dfrac{1}{4}$$ of its volume at $${27}^{o}C$$. If $$\gamma=1.4$$. The new temperature will be:

A
$$300\times {(4)}^{0.4}K$$

B
$$150\times {(4)}^{0.4}K$$

C
$$250\times {(4)}^{0.4}K$$

D
none of these

Solution

For adiabatic change, the equation is
$$T{V}^{\gamma-1}=$$ constant
$${T}_{1}{V}_{1}^{\gamma-1}-{T}_{2}{V}_{2}^{\gamma-1}$$
$$(27+273){V}_{1}^{\gamma-1}={T}_{2}{ \left( \cfrac { { V }_{ 1 } }{ { 4 } } \right) }^{ \gamma -1 }$$
$$300\times {V}_{1}^{\gamma-1}=\cfrac { { T }_{ 2 }\times { V }_{ 1 }^{ \gamma -1 } }{ { 4 }^{ \gamma -1 } } \Rightarrow { T }_{ 2 }=300\times { 4 }^{ \gamma -1 }$$
$${ T }_{ 2 }=300\times { 4 }^{ 1.4-1 }=300\times { 4 }^{ 0.4 }K$$

Hence, option A is correct
Question 2
1 / -0

Therm is the unit of

A
heat

B
temperature

C
thermometry

D
work

Solution

Therm is a non SI unit of heat energy equal to ICCF of natural gas and 10000 B+u { British thermal units}.
Hence, OPTION (A) heat.
Question 3
1 / -0

Combustion is

A
exothermic reaction

B
endothermic reaction

C
addition reaction

D
None of these

Solution

$$\textbf{Explanation:}$$
$$\bullet$$Combustion is an oxidation reaction and during this reaction, heat is produced. In this type of reaction, bonds are broken first and then new bonds may be created results in new material formation. So, it is always an exothermic reaction.

$$Answer:$$

Hence, option A is the correct option.
Question 4
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Specific heat of a gas undergoing adiabatic change is:

A
zero

B
infinite

C
positive

D
negative

Solution

For an adiabatic change, P, V and T changes. No heat exchange takes place.
$$\therefore dQ=0$$
$$\therefore n.c.d T = 0$$
$$\therefore $$ Specific heat, c = 0
Question 5
1 / -0

By opening the door of a refrigerator which is inside a room, the temperature of room ____________.

A
first decreases then increases

B
remains unchanged

C
increases

D
decreases

Solution

When the door of a refrigerator is opened the temperature of the room increases as the room is acting as the source here where the heat is rejected.
Question 6
1 / -0

The second law of thermodynamics says that in a cyclic process

A
Work cannot be converted into heat

B
Heat cannot be converted into work

C
Work cannot be completely converted into heat

D
Heat cannot be completely converted into work

Solution

Second law of thermodynamics states that in a cyclic process heat cannot be converted completely into work because attainment of $$0K$$ temperature is impossible.
Question 7
1 / -0

Which of the following thermodynamic relation is correct?

A
$$dG=Vdp-SdT$$

B
$$dE=pdV+TdS$$

C
$$dH=-Vdp+TdS$$

D
$$dG=Vdp+SdT$$

Solution

$$dG=dH-TdS-SdT$$ (as $$G=H-TS)$$
Again, $$H=E+pV$$
$$\therefore\;dH=dE+pdV+Vdp$$
and $$dE=Tds-pdV$$
Thus, $$dG=(TdS-pdV)+pdV+Vdp-TdS-SdT=Vdp-SdT$$
Question 8
1 / -0

Hess's law is based on:

A
law if conservation of mass

B
law of conservation of energy

C
first law of thermodynamics

D
None of the above

Solution

Hint: Regardless of how many stages or steps a reaction has, Hess's Law of Constant Heat Summation maintains.

Explanation:
Hess's law is actually recognized as an expression of the concept of energy conservation, which is also expressed in the first law of thermodynamics, as well as the fact that the enthalpy of a chemical process is independent of the path taken from the initial to the final state.
Hence, option C is correct.
Question 9
1 / -0

Which of the following is not a state function?

A
Work-done in adiabatic process

B
Work done in isothermal process

C
Heat at constant pressure

D
Heat at constant volume

Solution

Work done in adiabatic process = negative of change in internal energy (from first law of thermodynamics). So it is a state function.
For heat at constant pressure,
$$dQ = nCp\Delta T$$
This is path independent.
For work done in isothermal process,
$$dQ = dW$$. This is path dependent and hence not a state function.
Heat at constant volume is equal to change in internal energy (from first law of thermodynamics). So it is also a state function.
Question 10
1 / -0

Refrigerators and freezers works by:

A
Converting hot air to cold air

B
Keeping hot air out with cold air pressure

C
Removing heat from inside themselves

D
Blowing cold inside them

E
Producing cold air

Solution

A gas refrigerant or sometimes air is key to refrigeration in freezers.
The gas refrigerant goes into the compressor and the process of compression makes the gas hot. That hot gas moves through the system getting colder and colder. By the time it reaches the valve – and is pushed through the very small opening – it has turned into a cold mist.

As the mist goes through the coils under the freezer compartment, it starts to evaporate and turn back into a gas. This evaporation process takes the heat from the freezer compartment with it and as the refrigerant starts to warm up, it is sent back to the compressor to start the process all over again. Eventually the process is completed and there is no heat left in the freezer meaning that all objects in the freezer are frozen.

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Thermodynamics Test - 20

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Question 1

$$300\times {(4)}^{0.4}K$$

$$150\times {(4)}^{0.4}K$$

$$250\times {(4)}^{0.4}K$$

none of these

Solution

Question 2

heat

temperature

thermometry

work

Solution

Question 3

exothermic reaction

endothermic reaction

addition reaction

None of these

Solution

Question 4

zero

infinite

positive

negative

Solution

Question 5

first decreases then increases

remains unchanged

increases

decreases

Solution

Question 6

Work cannot be converted into heat

Heat cannot be converted into work

Work cannot be completely converted into heat

Heat cannot be completely converted into work

Solution

Question 7

$$dG=Vdp-SdT$$

$$dE=pdV+TdS$$

$$dH=-Vdp+TdS$$

$$dG=Vdp+SdT$$

Solution

Question 8

law if conservation of mass

law of conservation of energy

first law of thermodynamics

None of the above

Solution

Question 9

Work-done in adiabatic process

Work done in isothermal process

Heat at constant pressure

Heat at constant volume

Solution

Question 10

Converting hot air to cold air

Keeping hot air out with cold air pressure

Removing heat from inside themselves

Blowing cold inside them

Producing cold air

Solution

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