Thermodynamics Test

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Thermodynamics Test - 13

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Question 1
1 / -0

In the isothermal expansion of a gas :

A
the work done by gas is zero

B
heat is taken from the gas

C
heat is neither given nor taken from gas

D
internal energy stored in the gas remains constant

Solution

Internal energy is a function of temperature only.
Therefore, in case of isothermal process, it will remain constant.
Question 2
1 / -0

The efficiency of a heat engine :

A
is independent of the temperature of the source and the sink

B
is independent of the working substance

C
can be 100%

D
is not affected by the thermal capacity of the source or the sink

Solution

$$\textbf{Hint}$$: Efficieny of heat engine
$$\textbf{Step 1:Efficiency of Heat Engine}$$:
A heat engine involves a thermodynamic process that converts the heat supply in it into mechanical work.
The efficiency $$\eta$$ of the heat engine is the ratio of the work to the heat supplied to the heat engine.
$$\eta=\dfrac{W}{Q_1}$$
Where $$W$$ is work done and $$Q_1$$ is the heat absorbed from the source.
$$\textbf{Step 2:Efficiency in terms of temperature}$$:
Efficiency is defined as work done per unit of heat consumed. It doesn't depend on the substance on which we work.
Efficiency $$\eta=\dfrac{Work output}{Heat Input}$$
$$\eta=\dfrac{T_1-T_2}{T_1}$$ where $$T_1$$ and $$T_2$$ are temperature of source and sink respectively.
$$\textbf{Step3:Conclusions}$$
$$\bullet$$Efficiency is dependent on the source and sink temperature. So, option A is not correct.
$$\bullet$$From the formula, we can say efficiency is independent of working substance.
Thus option B is correct.
$$\bullet$$Here efficiency will never be $$100%$$ because $$\dfrac{T_2}{T_1}$$ is always a positive number and less than one because sink temperature is always less than source temperature. Thus option C is incorrect.
$$\bullet$$The amount of heat taken from the source and heat going to dissipate on the sink depend on its thermal capacity. Thus option D is incorrect.
Question 3
1 / -0

Which of the following is true in the case of an adiabatic process where $$\gamma =C_{p}/C_{v}$$?

A
$$P^{1-\gamma }T^{\gamma }=constant$$

B
$$P^{\gamma }T^{1-\gamma}=constant$$

C
$$PT^{\gamma}=constant$$

D
$$P^{\gamma}T=constant$$

Solution

In an adiabatic process heat supplied is zero. Hence internal energy change equals work done.

Hence, $$W=\Delta U$$

$$-PdV=\dfrac { nR }{ \gamma -1 } dT=\dfrac { PdV+VdP }{ \gamma -1 } $$

$$-\gamma PdV=VdP$$

$$-\gamma \dfrac { dV }{ V } =\dfrac { dP }{ P } $$

$$ln(P)+\gamma ln(V)=C$$

$$P{ V }^{ \gamma }=constant$$

Putting $$V=\dfrac { nRT }{ P } $$

$${ P }^{ 1-\gamma }{ T }^{ \gamma }=constant$$

Answer is option A.
Question 4
1 / -0

A sink, that is the system where heat is rejected, is essential for the conversion of heat into work. From which law does the above inference follow?

A
Zeroth

B
First

C
Second

D
Third

Solution

A sink, that is the system where heat is rejected, is essential for the conversion of heat into work. Otherwise, change in entropy of the universe won't be positive.Which is must according to the second law of thermodynamics.
Question 5
1 / -0

Which of the following laws of thermodynamics leads to the inference that it is difficult to convert whole of heat into work :

A
zeroth

B
second

C
first

D
third

Solution

The second law of thermodynamics states that when energy change from one form to another form, entropy in a closed system increases.
In other words it can be stated as,
It is impossible to construct a device which produces no other effect than transfer of heat from lower temperature body to higher temperature body.
Hence it is clear that second law of thermodynamics leads to the interference that is difficult to convert whole of heat into work.
Question 6
1 / -0

When heat is added to a system, which of the following is not possible?

A
Internal energy of the system increases

B
Work is done by the system

C
Neither internal energy increases nor work is done by the system

D
Internal energy increases and also work is done by the system

Solution

From 1st law of thermodynamics,
$$\Delta Q = \Delta U + \Delta W$$
From the law, if heat is supplied to the system, there has to be an increase in it's internal energy or it has to do some work or both.
Question 7
1 / -0

For an adiabatic process the relation between V and T is given by

A
$$TV^{\gamma }=constant$$

B
$$T^{\gamma }V=constant$$

C
$$TV^{1-\gamma }=constant$$

D
$$TV^{\gamma-1 }=constant$$

Solution

For a gas going under an adiabatic process, we have $$T \times V^{\gamma -1}=k$$, where k is some constant.
Question 8
1 / -0

The second law of thermodynamics implies :

A
whole of heat can be converted into mechanical energy

B
no heat engine can be 100% efficient

C
every heat engine has an efficiency of 100%

D
a refrigerator can reduce the temperature to absolute zero

Solution

second law of thermodynamics states that ,
"It is impossible to construct a device which produces no other effect than transfer of heat from lower temperature body to higher temperature body."
$$\therefore$$ The second law of thermodynamics implies that no heat engine can be 100% efficient.
Question 9
1 / -0

$$dU + dW = 0$$ is valid for :

A
adiabatic process

B
isothermal process

C
isobaric process

D
isochoric process

Solution

According to first law of thermodynamics we have $$dQ = dU + dW$$
For an adiabatic process, we know that $$dQ=0$$
Therefore we have $$ dU + dW = 0$$
Question 10
1 / -0

The ratio of slopes of adiabatic and isothermal curves is :

A
$$\gamma $$

B
$$\dfrac{1}{\gamma} $$

C
$$\gamma ^{2} $$

D
$$\gamma ^{3} $$

Solution

Slope of PV graph for adiabatic is given by $$\gamma (\dfrac{P}{V})$$
and that for an isothermal process is given by $$ \dfrac{P}{V}$$
Hence their ratio is $$\gamma$$.

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Thermodynamics Test - 13

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Thermodynamics Test - 13

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Question 1

the work done by gas is zero

heat is taken from the gas

heat is neither given nor taken from gas

internal energy stored in the gas remains constant

Solution

Question 2

is independent of the temperature of the source and the sink

is independent of the working substance

can be 100%

is not affected by the thermal capacity of the source or the sink

Solution

Question 3

$$P^{1-\gamma }T^{\gamma }=constant$$

$$P^{\gamma }T^{1-\gamma}=constant$$

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

$$P^{\gamma}T=constant$$

Solution

Question 4

Zeroth

First

Second

Third

Solution

Question 5

zeroth

second

first

third

Solution

Question 6

Internal energy of the system increases

Work is done by the system

Neither internal energy increases nor work is done by the system

Internal energy increases and also work is done by the system

Solution

Question 7

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

$$T^{\gamma }V=constant$$

$$TV^{1-\gamma }=constant$$

$$TV^{\gamma-1 }=constant$$

Solution

Question 8

whole of heat can be converted into mechanical energy

no heat engine can be 100% efficient

every heat engine has an efficiency of 100%

a refrigerator can reduce the temperature to absolute zero

Solution

Question 9

adiabatic process

isothermal process

isobaric process

isochoric process

Solution

Question 10

$$\gamma $$

$$\dfrac{1}{\gamma} $$

$$\gamma ^{2} $$

$$\gamma ^{3} $$

Solution

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