Thermodynamics Test

Result

Thermodynamics Test - 52

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

Consider a cycle followed by an engine as shown in figure
1 to 2 is isothermal
2 to 3 is adiabatic
3 to 1 is adiabatic
Such a process does not exist because

A
heat is completely converted to mechanical energy in such a process, which is not possible.

B
mechanical energy is completely converted to heat in this process, which is not possible.

C
curves representing two adiabatic processes can intersect.

D
Curves representing an adiabatic process and an isothermal process dont intersect.

Solution

Heat change is converted to mechanical energy. Since efficiency of no machine is $$100\%$$ it is not possible.
Question 2
1 / -0

Two moles of helium gas undergo a cyclic process as shown in figure. Assuming the gas to be ideal. The net work done by the gas is:

A
200 R In 2

B
100 R In 2

C
300 R In 2

D
400 R In 2

Solution

Process $$AB$$ and $$CD$$ $$-$$ isobaric
Workdone $$=P\Delta V=nR\Delta T$$
$$WAB=nR\left( 100 \right) $$
$$WCD=-nR\left( 100 \right) $$
Process $$BC$$ and $$DA$$ $$\Rightarrow$$ isothermal
Work done $$=nRTln\dfrac { { P }_{ 1 } }{ { P }_{ 2 } } $$
$$WBC=nRTln\dfrac { 1 }{ 2 } =-400nRln\dfrac { 1 }{ 2 } $$
$$W\Delta A=nRTln2=300nRln2$$
Total work done $$=100nR-100nR+400Rln2-300nRln2$$
$$=100nRln2$$
$$=200Rln2$$ $$(n=2)$$
Question 3
1 / -0

Some gas $$(\dfrac { { C }_{ p } }{ { C }_{ v } }$$ = $$\gamma$$ = 1.25) follows the cycle ABCDA as shown in the figure. The ratio of the energy given out by the gas to its surroundings during the isochoric section of the cycle to expansion work done during the isobaric section of the cycle is:

A
2

B
4

C
6

D
8

Solution

Let $$n$$ moles of gas follows the cycle $$ABCDA$$
$$Q_{\text{isoch or ic}}=nC_v\Delta T=\dfrac{nR}{\gamma-1}(T_B-T_C)$$

$$W_{isobaric}=nR\Delta T=nR(T_A-T_D)=nR(T_{B-T_C})$$

$$\therefore$$ Required ratio $$=\dfrac{Q_{\text{isoch or ic}}}{W_{isobaric}}=\dfrac{1}{\gamma-1}=4$$
Question 4
1 / -0

One mole of an ideal gas goes from an initial state A to final state B via two processes : It first undergoes isothermal expansion from volume V to 3V and then its volume is reduced from 3V to V at constant pressure. The correct P-V diagram representing the two processes is:

A

B

C

D

Solution

Given that gas A$$\rightarrow$$B
firstly isothermal expansion$$=$$V to $$3V$$
then volume reduced from $$3V$$ to V at constant pressure
(I) For isothermal expansion $$T=$$constant
$$\Rightarrow PV=nRT=$$constant
$$\Rightarrow PV=$$constant
Hence hyperbolic curve
(II) Now for isobaric compression $$P=$$constant
$$PV=nRT$$
$$=KV=nRT=$$straight line is correct.
Question 5
1 / -0

The freezer in a refrigerator is located at the top section so that;

A
The entire chamber of the refrigerator is cooled quickly due to convection

B
The motor is not heated

C
The heat gained from the environment is high

D
The heat gained from the environment is low

Solution

The freezer in a refrigerator is located at the top section so that the entire chamber of the refrigerator is cooled quickly due to convection.
The correct option is A.
Question 6
1 / -0

NA heat engine has an efficiency n.Temperatures of source and sink are each decreased by 100 K. The efficiency of the engine:

A
increases

B
decreases

C
remains constant

D
becomes 1

Solution

Efficiency of heat engine,
$$n=1-\dfrac{T_2}{T_1}=\dfrac{T_1-T_2}{T_1}$$. . . . . .(1)
According to question,
$$T_1\rightarrow T_1-100$$
$$T_2\rightarrow T_2-100$$
New efficiency, $$n'=1-\dfrac{T_2-100}{T_1-100}$$
$$n'=\dfrac{T_1-T_2}{T_1-100}$$
From equation (1) and (2), we can conclude that
$$n'\propto \dfrac{1}{T_1-100}$$
The efficiency of the engine is increases.
The correct option is A.
Question 7
1 / -0

An engine has an efficiency of $$0.25$$ when the temperature of sink is reduced by $$58C$$, if its efficiency is doubled, then the temperature of the source is:

A
$$150K$$

B
$$222K$$

C
$$242K$$

D
$$232K$$

Solution

Let us consider $$T_1=$$ temperature of the source
$$T_2=$$ temperature of the sink
The efficiency of the engine,
$$0.25=1-\dfrac{T_2}{T_1}$$
$$\dfrac{T_2}{T_1}=0.75$$
$$T_2=0.75T_1$$. . . . . . . . .(1)
When the temperature of the sink is reduced by $$58^0C$$, then the efficiency is double,
$$2\times 0.25=1-\dfrac{T_2-58}{T_1}$$
$$\dfrac{T_2-58}{T_1}=0.5$$
$$T_2-58=0.5T_1$$
$$0.75T_1-58=0.5T_1$$ .........(from equation 1)
$$0.25T_1-58=0$$
$$0.25T_1=58$$
$$T_1=\dfrac{58}{0.25}$$
$$T_1=232 K$$
The correct option is D.
Question 8
1 / -0

In a heat engine, the temperature of the source and sink is $$500 K$$ and $$375 K$$. If the engine consumes $$ 25\times{ 10 }^{ 5 }$$ per cycle, the work is done per cycle is:

A
$$6.25\times{ 10 }^{ 5 }J$$

B
$$3\times{ 10 }^{ 5 }J$$

C
$$2.9\times{ 10 }^{ 5 }J$$

D
$$4\times{ 10 }^{ 4 }J$$

Solution

Given,
$$T_1=500K$$
$$T_2=375K$$
$$Q_1=25\times 10^5J$$

Efficiency of the engine,
$$\eta=1-\dfrac{T_2}{T_1}$$

$$\eta=1-\dfrac{375}{500}$$

$$\eta=1-0.75$$
$$\eta=0.25=25$$%

We know that, $$\eta=\dfrac{W}{Q_1}$$

$$W=\eta Q_1$$
$$W=0.25\times 25\times 10^5$$
$$W=6.25\times 10^5J$$
The correct option is A.
Question 9
1 / -0

A one mole of an ideal gas expands adiabatically at constant pressure such that its temperature T$$\propto \dfrac { 1 }{ \sqrt { V }}$$. The value of the adiabatic constant of gas is:

A
1.3

B
1.5

C
q1.67

D
2.0

Solution

In adiabatic process,
$$PV^{\gamma}=k(constant)$$. . . . . .(1)
From ideal gas equation,
$$PV=nRT$$
$$P=\dfrac{nRT}{V}$$
From equation (1),
$$\dfrac{2RT}{V}V^{\gamma}=k$$
$$TV^{\gamma-1}=k'(constant)$$. . . . . .(2)
Given,
$$T\propto \dfrac{1}{\sqrt{V}}$$
$$TV^{1/2}=constant$$. . . . . . .(3)
By comparing equation (2) and (3), we get
$$\gamma-1=\dfrac{1}{2}=0.5$$
The value of adiabatic constant of gas is $$1.5$$.
$$\gamma=1+0.5=1.5$$
The correct option is B.
Question 10
1 / -0

Two bars of same length and same cross -sectional area but of different thermal conductivities $$K_1$$ and $$K_2$$ are joined end to end as shown in the figure .One end of the compound bar is at temperature $$T_1$$ and the opposite end at temperature $$T_2$$ $$(where T_1 > T_2)$$.
The temperature of the junction is

A
$$\dfrac{K_1T_1+K_2T_2}{K_1+K_2}$$

B
$$\dfrac{K_1T_2+K_2T_1}{K_1+K_2}$$

C
$$\dfrac{K_1(T_1+T_2)}{K_2}$$

D
$$\dfrac{K_2(T_1+T_2)}{K_1}$$

Solution

Let $$L$$ and $$A$$ be length and area of cross-section of each bar respectively.

$$\therefore $$ Heat current through the bar 1 is
$$H_1=\dfrac{K_1A(T_1-T_0)}{L}$$

At steady state,$$H_1=H_2$$

$$\therefore \dfrac{K_1A(T_1-T_0)}{L}=\dfrac{K_2A(T_0-T_2)}{L}$$

$$K_1(T_1-T_0)=K_2(T_0-T_2)$$

$$K_1T_1-K_1T_0=K_2T_0-K_2T_2$$

$$K_1T_0+K_2T_0=K_1T_1+K_2T_2$$

$$T_0(K_1+K_2)=K_1T_1+K_2T_2$$

$$T_0=\dfrac{K_1T_1+K_2T_2}{(K_1+K_2)}$$