Thermodynamics Test

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Thermodynamics Test - 73

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Weekly Quiz Competition

Question 1
1 / -0

The molar specific heat at constant pressure for a monoatomic gas is

A
$$ \dfrac{3}{2}R$$

B
$$ \dfrac{5}{2}R$$

C
$$ \dfrac{7}{2}R$$

D
$$4R$$

Solution
Question 2
1 / -0

Which of the following is an example of the first law of thermodynamics ?

A
The specific heat of an object explains how easily it changes temperatures.

B
While melting, an ice cube remains at the same temperature.

C
When a refrigerator is unplugged, everything inside of it returns to room temperature after some time.

D
After falling down the hill, a ball's kinetic energy plus heat energy equals the initial potential energy.

Solution

Explanation:
The first law of thermodynamics is another wording of the law of conservation of energy. Effectively it states that energy cannot be created or destroyed, but it can change forms.
This means that, in the given situation of the ball rolling down the hill, the total initial energy equals the final kinetic energy plus heat.
The zeroth law of thermodynamics states that if a system is in equilibrium with two other systems, then the two other systems are in equilibrium with each other.
The second law of thermodynamics states the entropy of a closed system will always increase.
The third law of thermodynamics states that absolute zero is the temperature at which entropy is zero.
Question 3
1 / -0

If a gas undergoes an isobaric process, which of the fol-lowing statements is true?

A
The temperature of the gas doesnt change.

B
Work is done on or by the gas.

C
No energy is transferred by heat to or from the gas.

D
The volume of the gas remains the same.

E
The pressure of the gas decreases uniformly.

Solution

In an isobaric process on an ideal gas, pressure is constant while the gas either expands or is compressed. Since the volume of the gas is changing, work is done either on or by the gas. Also, from the ideal gas law with pressure constant, $$P\Delta V = nR\Delta T$$; thus, the gas must undergo a change in temperature having the same sign as the change in volume. If $$\Delta V >0$$, then both $$\Delta T$$ and the change in the internal energy of the gas are positive $$(\Delta U > 0)$$. However, when $$\Delta V >0$$, the work done on the gas is negative $$( \Delta W < 0)$$, and the first law of thermodynamics says that there must be a positive transfer of energy by heat to the gas $$(Q = \Delta U - W> 0)$$. When $$\Delta V <0$$, a similar argument shows that $$\Delta U <0, W >0$$, and $$Q = \Delta y - W < 0$$. Thus, all of the other listed choices are false statements.
Question 4
1 / -0

An engine does 15.0 kJ of work while exhausting 37.0 kJ to a cold reservoir. What is the efficiency of the engine?

A
0.150

B
0.288

C
0.333

D
0.450

E
1.20

Solution

Efficiency of system :

$$e=\dfrac{W}{Q_{in}}$$

Here $$W=15\,kJ$$ and $$Q=37\,kJ$$

putting values:

$$e=\dfrac{15}{37}=0.45$$

Hence efficiency of the system is 0.45

Option D is correct
Question 5
1 / -0

Of the following, which is not a statement of the second law of thermodynamics?

A
No heat engine operating in a cycle can absorb energy from a reservoir and use it entirely to do work.

B
No real engine operating between two energy reservoirs can be more efficient than a Carnot engine operating between the same two
reservoirs.

C
When a system undergoes a change in state, the change in the internal energy of the system is the sum oaf the energy transferred to the system by heat and the work done on the system.

D
The entropy of the Universe increases in all natural processes.

E
Energy will not spontaneously transfer by heat from a cold object to a hot object.

Solution

First law of thermodynamics state When a system undergoes a change in state, the change in the internal energy of the system is the sum oaf the energy transferred to the system by heat and the work done on the system.
This is not a statement of second law of thermodynamics.
Hence option C is correct.
Question 6
1 / -0

A thermally insulated rigid container contains an ideal gas. It is heated through a resistance coil of 100$$\Omega $$ by passing a current of 1 A for five minutes, then change in internal energy of the gas is

A
0 KJ

B
30 KJ

C
10 KJ

D
20 KJ

Solution

Heat H supplied to the gas in container: $$H = i^2 \times R\times dt$$
where is is the current in the resistor( 1A), R is the resistance( 100 $$\Omega $$ and dt is the duration of time( 5 mins= 300 secs)
$$ \therefore H= 1^2 \times 100 \times 300 =30000J= 30 \ kJ$$
Since the container is rigid, no work is done when the the gas expands. $$dW=0$$ in the equation $$dQ=dU+dW$$
As the work done is 0 so $$dQ=dU$$
Then $$\textbf{The change in internal energy is equal to =30kJ}$$
Question 7
1 / -0

Two samples $$A$$ and $$B$$ of a gas initially of same pressure and temperature are compressed from a volume $$V$$ to a volume $$\displaystyle \dfrac{\mathrm{V}}{2}$$ such that $$A$$ is compressed isothermally while $$B$$ is compressed adiabatically. The final

A
pressure of $$A$$ is greater than that of $$B$$

B
pressure of $$A$$ is equal to that of $$B$$

C
pressure of $$A$$ is less than that of $$B$$

D
pressure of $$A$$ is twice the pressure of $$B$$

Solution

Ideal gas equation: $$PV=RT$$

Isothermal process:
Since temp T is constant $$P_1V_1=P_2V_2$$
$$V_1=V$$ as given
$$V_2=\frac {V}{2}$$ as given

$$ \therefore P_1 \times V=P_2 \times \dfrac{V}{2}$$

$$ \therefore P_2=2P_1$$

Adiabatic process:

$$V_1=V$$ as given
$$V_2=\dfrac {V}{2}$$ as given

$$PV^{\gamma}=$$constant

$$P_1{V_1}^{\gamma}=P_2{V_2}^{\gamma}$$

$$ \therefore P_1V^{\gamma}=P_2(\dfrac{V}{2})^{\gamma}$$

$$ \therefore P_2=P_1 \times 2^{\gamma} $$

Since $$\gamma$$ > 1, $$P_2$$ > $$P_1$$

Hence, $$(P_2)_{isothermal}$$ < $$(P_2)_{adiabatic}$$
Question 8
1 / -0

A monatomic gas is suddenly compressed to 1/8th of its initial volume adiabatically. The ratio of its final pressure to the initial pressure is ($$\gamma = $$ 5/3)

A
32

B
16

C
2

D
64

Solution

In an adiabatic process $$PV^{\gamma}=constant$$
$$ \therefore P_1V_1^{\gamma}=P_2V_2^{\gamma}$$ .....(1)
Given $$\gamma =\dfrac{5}{3}$$
$$V_2=\dfrac{V_1}{8}$$
substituting $$V_2$$ in eqn(1) $$ P_1V_1^{\dfrac {5}{3}}= P_2(\dfrac{V_1}{8})^{\dfrac{5}{3}}$$
$$ \Rightarrow P_2=32P_1$$
Question 9
1 / -0

An ideal gas at $$27^oC$$ is compressed adiabatically to $$\dfrac{8}{27}$$ of its initial value. If $$\gamma= \dfrac{5}{3}$$, find the rise in temperature.

A
$$275^oC$$

B
$$375^oC$$

C
$$400^oC$$

D
$$300^oC$$

Solution

For an adiabatic process $$TV^{\gamma -1}=constant$$
For the states having temp and volume $$(T_1, V_1)$$ and $$(T_2, V_2)$$
$$T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$$ .....(1)
Given:
$$V_2=\dfrac{8}{27}V_1$$
$$T_1=27^oC=27+273=300K$$
$$\gamma = \dfrac{5}{3}$$
Substituting the values in eqn(1)
$$300 \times V_1^{\dfrac {5}{3} -1} = T_2 \times ( \dfrac{8}{27}V_1)^{\dfrac {5}{3} -1}$$
$$ \Rightarrow 300= T_2\times (\dfrac{2}{3})^{3 \times \dfrac{2}{3}}=675^oK$$
Rise in temperature = 675 - 300 = $$375^{\circ} C$$
Question 10
1 / -0

A diatomic ideal gas is compressed adiabatically in order to increase the pressure by $$3.5$$%. The percentage increment in temperature of the gas is approximately

A
$$1$$

B
$$1.4$$

C
$$2$$

D
$$2.8$$

Solution

In an adiabatic process,
$$\dfrac {T^{\gamma}}{P^{\gamma -1}}=Const$$
$$ \Rightarrow T^{\gamma}= Const \times P^{\gamma -1}$$ ....(1)
Taking differential of both sides,

$$\gamma T^{\gamma -1} dT= Const \times (\gamma -1)P^{\gamma-2}dP$$

$$\gamma T^{\gamma} \dfrac{dT}{T} = Const \times (\gamma -1)P^{\gamma-1}\dfrac{dP}{P}$$

$$ \Rightarrow \gamma T^{\gamma} \dfrac{dT}{T} = \dfrac {T^{\gamma}}{P^{\gamma -1}}\times (\gamma -1)P^{\gamma-1}\frac{dP}{P}$$

$$ \Rightarrow \dfrac{dT}{T} = \dfrac {\gamma-1}{\gamma} * \dfrac{dP}{P}$$

In an adiabatic process,

$$\dfrac {T^{\gamma}}{P^{\gamma -1}}=Const$$

$$ \Rightarrow T^{\gamma}= Const \times P^{\gamma -1}$$ ....(1)

Taking differential of both sides,
$$\gamma T^{\gamma -1} dT= Const \times (\gamma -1)P^{\gamma-2}dP$$

$$\gamma T^{\gamma} \dfrac{dT}{T} = Const \times (\gamma -1)P^{\gamma-1}\dfrac{dP}{P}$$

$$ \Rightarrow \gamma T^{\gamma} \dfrac{dT}{T} = \dfrac {T^{\gamma}}{P^{\gamma -1}}\times (\gamma -1)P^{\gamma-1}\dfrac{dP}{P}$$

$$ \Rightarrow \dfrac{dT}{T} = \dfrac {\gamma-1}{\gamma} \times \dfrac{dP}{P}$$ .....(1)

$$\gamma = \dfrac{2+f}{f}$$ where f is the no of degrees of freedom of the gas molecule
A di-atomic molecule has 3 translational and 2 rotational degrees of freedom. Hence $$f=3+2=5$$
$$ \Rightarrow \gamma = \dfrac {2+f}{f}=\dfrac{2+5}{5}=\dfrac{7}{5}$$
Given $$ \dfrac {dP}{P}=0.035$$

substituting $$\dfrac {dP}{P}$$ and $$\gamma$$ in eqn(1)

$$ \dfrac{dT}{T}=\dfrac {(\dfrac{7}{5}-1)}{\dfrac{7}{5}}\times 0.035 \times 100=1$$

Thus, the percentage increment in temp of the gas is 1%

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Thermodynamics Test - 73

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Question 1

$$ \dfrac{3}{2}R$$

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

$$ \dfrac{7}{2}R$$

$$4R$$

Solution

Question 2

The specific heat of an object explains how easily it changes temperatures.

While melting, an ice cube remains at the same temperature.

When a refrigerator is unplugged, everything inside of it returns to room temperature after some time.

After falling down the hill, a ball's kinetic energy plus heat energy equals the initial potential energy.

Solution

Question 3

The temperature of the gas doesnt change.

Work is done on or by the gas.

No energy is transferred by heat to or from the gas.

The volume of the gas remains the same.

The pressure of the gas decreases uniformly.

Solution

Question 4

0.150

0.288

0.333

0.450

1.20

Solution

Question 5

No heat engine operating in a cycle can absorb energy from a reservoir and use it entirely to do work.

No real engine operating between two energy reservoirs can be more efficient than a Carnot engine operating between the same tworeservoirs.

When a system undergoes a change in state, the change in the internal energy of the system is the sum oaf the energy transferred to the system by heat and the work done on the system.

The entropy of the Universe increases in all natural processes.

Energy will not spontaneously transfer by heat from a cold object to a hot object.

Solution

Question 6

0 KJ

30 KJ

10 KJ

20 KJ

Solution

Question 7

pressure of $$A$$ is greater than that of $$B$$

pressure of $$A$$ is equal to that of $$B$$

pressure of $$A$$ is less than that of $$B$$

pressure of $$A$$ is twice the pressure of $$B$$

Solution

Question 8

32

16

2

64

Solution

Question 9

$$275^oC$$

$$375^oC$$

$$400^oC$$

$$300^oC$$

Solution

Question 10

$$1$$

$$1.4$$

$$2$$

$$2.8$$

Solution

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No real engine operating between two energy reservoirs can be more efficient than a Carnot engine operating between the same two
reservoirs.