Thermodynamics Test

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Thermodynamics Test - 64

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Question 1
1 / -0

Find coefficient of performance, If in a mechanical refrigerator, the lower temperature coils of evaporator are $$-23^oC$$ compressed gas in condenser has a temperature of $$77^oC$$. The coefficient of performance is

A
$$70\%$$

B
$$20\%$$

C
$$0.23\%$$

D
$$2.5\%$$

Solution
Question 2
1 / -0

In a thermodynamic system, $$Q$$ represents the energy transferred to or from a system by heat and $$W$$ represents the energy transferred to or from a system by work
$$Q > 0$$ and $$W=0$$
$$Q < 0$$ and $$W=0$$
$$W > 0$$ and $$Q=0$$
$$W > 0$$ and $$Q=0$$
Which of the above will lead to an increase in the internal energy of the system?

A
$$I\,only$$

B
$$II\,only$$

C
$$I\, and\,IV\,only$$

D
$$II\, and\,III\,only$$

E
$$II\, and\,IV\,only$$

Solution

According to first law of thermodynamics,
$$Q=\Delta U+W$$
Or $$\Delta U=Q-W$$
According to given conditions
I. $$Q > 0$$ and $$W=0$$
Or $$\Delta U=Q$$
Here $$Q > 0$$, therefore $$\Delta V$$ will be increase.

II. $$Q < 0$$ and $$W=0$$
Or $$\Delta U=Q$$
Here $$Q > 0$$, therefore $$\Delta U$$ will be decrease.

III. $$W > 0$$ and $$Q=0$$
Or $$\Delta U=-W$$
Here $$W > 0$$, therefore $$\Delta U$$ will be decrease.

IV. $$W < 0$$ and $$Q=0$$
Or $$\Delta U=-W$$
Here $$W < 0$$, therefore $$\Delta V$$ will be decrease.
Hence, condition I and IV will lead to an increase in the internal energy of the system. Therefore, option I and IV only is correct
Question 3
1 / -0

In an isobaric process, the work done by a di-atomic gas is $$10$$J, the heat given to the gas will be?

A
$$35J$$

B
$$30J$$

C
$$45J$$

D
$$60J$$

Solution

For constant pressure process
$$\dfrac{W}{Q}=\dfrac{nR\Delta T}{nC_p\Delta T}=\dfrac{nR\Delta T}{n\left(\dfrac{f}{2}+1\right)R\Delta T}=\dfrac{1}{f/2+1}$$

$$\dfrac{W}{Q}=\dfrac{1}{\left(\dfrac{5}{2}+1\right)}=\dfrac{2}{7}$$

$$\implies Q=\dfrac{7}{2}W=\dfrac{7}{2}\times 10=35J.$$
Question 4
1 / -0

Air in a sealed syringe is slowly compressed by moving the piston. The temperature of the air stays the same. Which statement about the air is correct?

A
The pressure of the air increases because its molecules now travel more slowly

B
The pressure of the air increases because the area of the syringe walls is now smaller.

C
The pressure of the air increases because its molecules now hit the syringe walls more frequently

D
The pressure of the air increases because its molecules now travel more quickly

Solution

Since the temperature of air remain constant.So, it is a isothermal process.
$$PV=K$$
This eq. shows that on increasing pressure ,the volume get reduced and vice-versa.
Since the syringe is slowly compressed , volume is decreasing .
Therefore the pressure inside the syringe will increases because the molecules now hit the syringe walls more frequently.
Question 5
1 / -0

5,6 L of the gas at STP is adiabatically compressed. to 0.7 L taking initial Temperature $$ T_1 $$ the W.D in the process is

A
9/8 RT

B
3/2 RT

C
15 / 8 RT

D
9 /2 RT

Solution
Question 6
1 / -0

Which of the following is a slow process?

A
Isothermal

B
Adiabatic

C
Isobaric

D
None of these

Solution
Question 7
1 / -0

A sample of an ideal gas is taken through the cyclic process $$abca$$ as shown in the figure. The change in the internal energy of the gas along the path $$ca$$ is - $$180J$$. The gas absorbs $$250 J$$ of heat along the path ab and $$60J$$ along the path $$bc$$. The work done by the gas along the path $$abc$$ is :

A
$$100\ J$$

B
$$120\ J$$

C
$$140\ J$$

D
$$130\ J$$
Question 8
1 / -0

An ideal gas is compressed adibatically to (8/27) of its original volume. if $$ \gamma = 5/3 $$ and rise in its temperature if $$ 375^0 C $$ , its initial temperature is

A
$$ 127^0 C $$

B
$$ 225^0 C$$

C
$$ 27^0 C $$

D
$$75^0 C $$

Solution
Question 9
1 / -0

Five moles of an ideal monatomic gas with an initial temperature of $$150^0C$$ expand and in the process absorb $$1500J$$ of heat and does $$2500J$$ of work. The final temperature of the gas in $$^0C$$ (ideal gas constant, $$R=8.314JK^{-1}mol^{-1}$$ ).

A
$$134$$

B
$$126$$

C
$$144$$

D
$$166$$

E
$$174$$

Solution

According to thermodynamics first law, $$\Delta Q=\Delta U+\Delta W$$
Given, that
$$\Delta Q=+1500J$$
$$\Delta W=+2500J$$
$$\therefore 1500-\Delta V+2500$$
$$\Rightarrow \Delta V=1000J$$
$$\Delta V=mC_V\Delta T$$............(1)
$$\because$$ For monoatomic gas,
$$C_V=\dfrac{3}{2}R=1.5\times 8.314$$
$$C_V=12.471$$
$$\therefore$$ From Equation (i)
$$\therefore -1000=5\times 12.471\times (T_2-T_1)$$
Or $$T_2=150-\dfrac{1000}{5\times 12.471}=134^0C$$
Question 10
1 / -0

Two identical systems, with heat capacity at constant volume that varies as $$C_v=bT^3$$ (where $$b$$ is a constant) are thermally isolated. Initially, one system is at a temperature $$100K$$ and the other is at $$200K$$. The systems are then brought to thermal contact and the combined system is allowed to reach thermal equilibrium. The final temperature (in K) of the combined system will be

A
$$171$$

B
$$141$$

C
$$150$$

D
$$180$$

E
$$125$$

Solution

Energy given by a system = Energy taken by another system
$$dQ_1=dQ_2$$
$$\displaystyle \Rightarrow \int^T_{100} mc_vdT=-\int^T_{200} mc_vdT$$

$$\displaystyle \Rightarrow \int^T_{100} bT^3 dT=-\int^T_{200} bT^3dT\Rightarrow (T^4)^T_{10}=-(T^4)^T_{200}$$

$$\Rightarrow T^4-100^4=-(T^4-200^4)$$
$$\Rightarrow 2T^4=17\times 10^8\Rightarrow T=171K$$

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Thermodynamics Test - 64

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Thermodynamics Test - 64

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Question 1

$$70\%$$

$$20\%$$

$$0.23\%$$

$$2.5\%$$

Solution

Question 2

$$I\,only$$

$$II\,only$$

$$I\, and\,IV\,only$$

$$II\, and\,III\,only$$

$$II\, and\,IV\,only$$

Solution

Question 3

$$35J$$

$$30J$$

$$45J$$

$$60J$$

Solution

Question 4

The pressure of the air increases because its molecules now travel more slowly

The pressure of the air increases because the area of the syringe walls is now smaller.

The pressure of the air increases because its molecules now hit the syringe walls more frequently

The pressure of the air increases because its molecules now travel more quickly

Solution

Question 5

9/8 RT

3/2 RT

15 / 8 RT

9 /2 RT

Solution

Question 6

Isothermal

Adiabatic

Isobaric

None of these

Solution

Question 7

$$100\ J$$

$$120\ J$$

$$140\ J$$

$$130\ J$$

Question 8

$$ 127^0 C $$

$$ 225^0 C$$

$$ 27^0 C $$

$$75^0 C $$

Solution

Question 9

$$134$$

$$126$$

$$144$$

$$166$$

$$174$$

Solution

Question 10

$$171$$

$$141$$

$$150$$

$$180$$

$$125$$

Solution

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