Thermodynamics Test

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Thermodynamics Test - 10

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Weekly Quiz Competition

Question 1
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Directions For Questions

An ideal gas is undergoing a cyclic thermodynamic process in different ways as shown in the corresponding $$P- V$$ diagrams in column 3 of the table. Consider only the path from state 1 to state 2. W denotes the corresponding work done on the system. The equations and plots in the table have standard notations as used in thermodynamic process. Here $$\gamma$$ is the ratio of heat capacities at constant pressure and constant volume. The number of moles in the gas is $$n$$.

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Which one of the following options correctly represents a thermodynamic process that is used as a correction in the determination of the speed of sound in an ideal gas?

A
$$(I)(ii)(Q)$$

B
$$(I)(iv)(Q)$$

C
$$(IV)(ii)(R)$$

D
$$(III)(iv)(R)$$

Solution

I: Work done by adiabatic process:

$$W_{1-2}= \dfrac{(P_2V_2 - P_1V_1}{\gamma -1})$$

II: Work done by isobaric process:
$$W_{1-2}= - PV_2+ PV_1 = P \Delta V$$

III : Work done by isochoric Process :

$$W_ {1- 2} = 0 $$. As the volume is constant throughout the process.

IV: Work done by isothermal :
$$W_{1-2} = - nRT \ \ ln {\dfrac{V_2}{V_1}}$$
Question 2
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A diatomic ideal gas is compressed adiabatically to $$\displaystyle \frac{1}{32}$$ of its initial volume. If the initial temperature of the gas is $$\mathrm{T}_{\mathrm{i}}$$ (in Kelvin) and the final temperature is $$\mathrm{a}\mathrm{T}_{\mathrm{i}}$$, the value of a is

A
4

B
5

C
2

D
3

Solution

For adiabatic process, we can write $$\mathrm{T}\mathrm{V}^{\gamma-1}=$$ constant
$$\gamma = \dfrac{7}{5}$$ for diatomic gases

$$\mathrm{T_{i}}\mathrm{V}^{\dfrac{7}{5}-1}=$$ k(constant) -------initial condition

$$\mathrm{aT_{i}}\mathrm{V}^{\dfrac{7}{5}-1}=$$ k(constant) ------final condition

$$\displaystyle \mathrm{T}\mathrm{V}^{\dfrac{7}{5}-1}=\mathrm{a}\mathrm{T}(\frac{\mathrm{V}}{32})^{\dfrac{7}{5}-1}$$
$$\therefore \mathrm{a}=4$$
Question 3
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A monoatomic gas at pressure $$P_1$$ and volume $$V_1$$ is compressed adiabatically to $$\displaystyle \frac{1}{8}$$th of its original volume. What is the final pressure of the gas

A
$$64P_1$$

B
$$P_1$$

C
$$16P_1$$

D
$$32P_1$$

Solution

Ideal gas equation, for an adiabatic process is
$$PV^{\gamma} =$$ constant or $$P_1V_1^{\gamma} = P_2V_2^{\gamma}$$
For monoatomic gas $$\displaystyle \gamma = \dfrac{5}{3}$$
$$\displaystyle P_1V_1^{5/3} = P_2 \left( \dfrac{V_1}{8} \right)^{5/3}$$
$$\Rightarrow P_2 = P_1 \times (2)^5 = 32 P_1$$
Question 4
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A monoatomic gas at a pressure $$P$$, having a volume $$V$$ expands isothermally to a volume $$2V$$ and then adiabatically to a volume $$16V$$. The final pressure of the gas is : (take $$\gamma = 5/3$$)

A
$$64 P$$

B
$$32 P$$

C
$$\dfrac{P}{64}$$

D
$$16P$$

Solution

Isothermally $$PV=P_12V$$
$$P_1=\dfrac {P}{2}$$
Adiabatically, $$\dfrac {P}{2}(2V)^{\gamma}=P_1(16V)^{\gamma}$$
$$P_f=\dfrac {P}{2}\left (\dfrac {1}{2^3}\right )^{\dfrac {5}{3}}$$
$$P_t=\dfrac {P}{2}\left (\dfrac {1}{2^3}\right )^{\dfrac {5}{3}}=\dfrac {P}{(2)(2)^5}$$
$$P_f=\dfrac {P}{6}$$
Question 5
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The volume(V) of a monatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is?

A
$$\displaystyle\frac{1}{3}$$

B
$$\displaystyle\frac{2}{5}$$

C
$$\displaystyle\frac{2}{7}$$

D
$$\displaystyle\frac{2}{3}$$

Solution

$$\dfrac{work \ done}{heat \ absorbed} = \dfrac{n C_p \Delta T - n C_v \Delta T}{ n C_p \Delta T} =1- \dfrac {1}{\gamma} = \dfrac{2}{5}$$
Question 6
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A mass of diatomic gas $$\left(\gamma = 1.4\right)$$ at a pressure of $$2\ atm$$ is compressed adiabatically so that its temperature rise from $$27^oC $$ to $$927^oC$$. The pressure of the gas is final state is

A
$$28\ atm$$

B
$$68.7\ atm$$

C
$$256\ atm$$

D
$$8\ atm$$

Solution

$$P{V}^{\gamma} =$$ constant $${T}_{1} = 273 + 27 = 300 K$$
$$P{\left(\displaystyle\frac{T}{P}\right)}^{\gamma} =$$ constant $${T}_{2} = 273 + 927 = 1200 K$$

$${P}^{1-\gamma}{T}^{\gamma} =$$ constant
$$\Rightarrow {P}_{1}^{1-\gamma}{T}_{1}^{\gamma} = {P}_{2}^{1-\gamma}{T}_{2}^{\gamma}$$
$$\Rightarrow {2}^{1-1.4} {\left(300\right)}^{1.4} = {P}_{2}^{1-1.4} \cdot {\left(1200\right)}^{1.4}$$
$$\Rightarrow { \left( \displaystyle\dfrac { { P }_{ 2 } }{ { P }_{ 1 } } \right) }^{ 1-\gamma } = { \left( \displaystyle\dfrac { { T }_{ 1 } }{ { T }_{ 2 } } \right) }^{ \gamma } \Rightarrow \displaystyle\dfrac { { P }_{ 2 } }{ { P }_{ 1 } } = { \left( \displaystyle\dfrac { { T }_{ 1 } }{ { T }_{ 2 } } \right) }^{ \displaystyle\dfrac { \gamma }{ 1-\gamma } }$$

$${ \left( \displaystyle\dfrac { { P }_{ 1 } }{ { P }_{ 2 } } \right) }^{ 1-\gamma } = { \left( \displaystyle\dfrac { { T }_{ 2 } }{ { T }_{ 1 } } \right) }^{ \gamma }$$

$${ \left( \displaystyle\dfrac { { P }_{ 1 } }{ { P }_{ 2 } } \right) }^{ 1-1.4 } = { \left( \displaystyle\dfrac { 1200 }{ 300 } \right) }^{ 1.4 }$$

$${ \left( \displaystyle\dfrac { { P }_{ 1 } }{ { P }_{ 2 } } \right) }^{ -0.4 }={ \left( 4 \right) }^{ 1.4 }$$

$${ \left( \displaystyle\dfrac { { P }_{ 2 } }{ { P }_{ 1 } } \right) }^{ 0.4 }={ 4 }^{ 1.4 }$$

$${ P }_{ 2 } = { P }_{ 1 }{ 4 }^{ \left( \displaystyle1.4/0.4 \right) } = { P }_{ 1 }{ 4 }^{ \left( \displaystyle 7 /2 \right) }$$

$$={ P }_{ 1 }\left( { 2 }^{ 7 } \right) =2\times 128=256$$
Question 7
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Two cylinders $$A$$ and $$B$$ of equal capacity are connected to each other via a stop cock. $$A$$ contains an ideal gas at standard temperature and pressure. $$B$$ is completely evacuated. The entire system is thermally insulated. The stop cock is suddenly opened. The process is :

A
adiabatic

B
isochoric

C
isobaric

D
isothermal

Solution

Hint: A thermodynamic process that happens without transmitting heat or mass between the thermodynamic system and its surroundings is known as an adiabatic process.

Correct Answer: Option A

Explanation of Correct Answer:
The entire system is entirely insulated, allowing for free gas expansion while maintaining a steady gas temperature.
As a result, the procedure will be adiabatic.
So, the correct answer is option (A).

Explanation of Incorrect Options:
There should be no change in volume for isochoric process, but there change in volume exist. So, Option (B) is incorrect.
There should be no change in pressure for isobaric process, but there change in pressure exist. So, Option (C) is incorrect.
There is exchange of Heat by maintaining constant Temperature in isothermal process, but there is no exchange in heat .So, Option (D) is incorrect.

Therefore, Option (B), Option (C) and Option (D) are incorrect. The correct answer is option (A).
Question 8
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Thermodynamic process are indicated in the following diagram.
Match the following:

Column - I Column - II
P. Process I a. Adiabatic
Q. Process II b. Isobaric
R. Process III c. Isochoric
S. Process IV d. Isothermal

A
P-d, Q- b, R-a, S-c

B
P-a, Q- c, R-d, S-b

C
P-c, Q- a, R-d, S-b

D
P-c, Q- d, R-b, S-a

Solution

Path (I) $$\rightarrow$$ constant volume $$\rightarrow$$ isochoric
Path (II) $$\rightarrow\, PV^r=k \,\rightarrow $$ adiabatic
Path (III) $$\rightarrow$$ constant temperature $$\rightarrow$$ Isothermal
Path (IV) $$\rightarrow$$ constant pressure $$\rightarrow$$ isobaric
Question 9
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For an adiabatic process

A
$$\displaystyle \Delta S=0$$

B
$$\displaystyle \Delta U=0$$

C
$$\displaystyle Q=0$$

D
$$\displaystyle W=0$$

Solution

Explanation:
In this process, the system is insulated from the surroundings and hence heat absorbed or released is zero. As no heat enters or leaves the system.
$$\Delta Q=0$$
Thus, the entropy also does not change, and $$\Delta S=0$$
Example: Assume we have a hot tea in Thermos flask. It will remain hot as there is no exchange of heat takes place because the walls of thermos is insulating.

Hence, option (A) is correct.
Question 10
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What is the slope for an isothermal process?

A
$$\dfrac { P }{ V } $$

B
$$-\dfrac { P }{ V } $$

C
Zero

D
$$\infty $$

Solution

For an isothermal process, $$PV=const$$
$$\Rightarrow PdV+VdP=0$$
$$\Rightarrow \dfrac { dP }{ dV } =\dfrac { -P }{ V } $$

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	Column - I		Column - II
P.	Process I	a.	Adiabatic
Q.	Process II	b.	Isobaric
R.	Process III	c.	Isochoric
S.	Process IV	d.	Isothermal

Thermodynamics Test - 10

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Thermodynamics Test - 10

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Question 1

$$(I)(ii)(Q)$$

$$(I)(iv)(Q)$$

$$(IV)(ii)(R)$$

$$(III)(iv)(R)$$

Solution

Question 2

4

5

2

3

Solution

Question 3

$$64P_1$$

$$P_1$$

$$16P_1$$

$$32P_1$$

Solution

Question 4

$$64 P$$

$$32 P$$

$$\dfrac{P}{64}$$

$$16P$$

Solution

Question 5

$$\displaystyle\frac{1}{3}$$

$$\displaystyle\frac{2}{5}$$

$$\displaystyle\frac{2}{7}$$

$$\displaystyle\frac{2}{3}$$

Solution

Question 6

$$28\ atm$$

$$68.7\ atm$$

$$256\ atm$$

$$8\ atm$$

Solution

Question 7

adiabatic

isochoric

isobaric

isothermal

Solution

Question 8

P-d, Q- b, R-a, S-c

P-a, Q- c, R-d, S-b

P-c, Q- a, R-d, S-b

P-c, Q- d, R-b, S-a

Solution

Question 9

$$\displaystyle \Delta S=0$$

$$\displaystyle \Delta U=0$$

$$\displaystyle Q=0$$

$$\displaystyle W=0$$

Solution

Question 10

$$\dfrac { P }{ V } $$

$$-\dfrac { P }{ V } $$

Zero

$$\infty $$

Solution

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