Thermodynamics Test

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Thermodynamics Test - 75

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Question 1
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Directions For Questions

A process in which no heat enters or leaves a system is called an adiabatic process, $$Q=0$$. This prevention of heat flow may be accomplished either by surrounding the system with a thick layer of heat- insulating material(such as cork, asbestos, fire brick or styrofoam), or by performing the process quickly. The flow of heat requires finite time. So any process performed quickly enough will be practically adiabatic. Applying the first law to an adiabatic process, we get.

$$U_2-U_1=\Delta U=-W$$(adiabatic process)
Thus, the change in the internal energy of a system, an adiabatic process is equal in magnitude to the work done by the system. If the work $$W$$ is negative, as when a system is compressed, then $$W$$ is positive, $$U_2$$ is greater than $$U_1$$ and the internal energy of the system increases. If $$W$$ is positive, as when system expands, the internal energy of the system decreases. An increase of internal energy is usually(but not always) accompanied by a rise in temperature, and a decrease in internal energy by a temperature drop.

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A tyre having volume $$0.02\ m^3$$ and pressure $$2.5\ atm$$ suddenly bursts. What is the new volume occupied by air present?

A
$$0.036\ m^3$$

B
$$0.063\ m^3$$

C
$$0.072\ m^3$$

D
none

Solution

$$P_1V^r_1=P_2V^r_2$$
or
$$V_2=(\displaystyle \frac{P_1V^r_1}{P_2})^\dfrac{1}{r}$$

$$=0.02 \times (\displaystyle \frac{2.5}{1})^{(\dfrac{1}{1.4})}=0.036\ m^3$$
Question 2
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Directions For Questions

One mole of an ideal gas is taken from an initial state $$P_o, V_o$$ and temperature $$T_o$$ through the following activities. ($$\gamma = \dfrac{C_p}{C_v}, C_p = \dfrac{7R}{2}, C_v = \dfrac{5R}{2}$$)
(i) Heating at constant volume to a temperature three times.
(ii) Adiabatic expansion to a volume $$2V_o$$
(iii) Cooling at constant volume to a temperature on third
(iv) Adiabatic compression so that it is returned to its initial state

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The pressure at point C is given by

A
$$\displaystyle{\dfrac{3P_0}{2^{0.4}}}$$

B
$$\displaystyle{\dfrac{3P_0}{2^{1.4}}}$$

C
$$3P_0\times 2^{0.4}$$

D
$$3P_0\times 2^{1.4}$$

Solution

Using ideal gas equation $$PV=nRT$$
Using equation at point A $$P_\circ{}V_\circ{}=nRT_\circ{}$$...(ii)
Now the temperature has been reduced to 1/3rd at constant volume to reach point N

Using equation at point B $$P_BV_\circ{}=nR{\dfrac{T_\circ{}}{3}} \Rightarrow P_B=3P_\circ{}$$
Now the gas is expanded to 3 times original volume through adiabatic process
$$PV^\gamma=constant$$

Using equation at point C $$P_B{V^\circ{}}^\gamma=P_C{2V_\circ{}}^\gamma\Rightarrow P_C=\dfrac{P_B}{2^\gamma}$$
$$P_B=3P_\circ{} \Rightarrow P_C=\dfrac{3P_\circ{}}{2^\gamma} ; \;gamma=1.4$$
$$P_C=\dfrac{3P_\circ{}}{2^{1.4}}$$
Hence correct answer is B
Question 3
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Directions For Questions

The work done during adiabatic reversible process is given by $$\displaystyle w=\dfrac { nR[{ T }_{ f }-{ T }_{ i }] }{ \gamma -1 } $$. Thus, if $$\displaystyle { T }_{ f }>{ T }_{ i }$$, then since $$\displaystyle \gamma-1=+ve\ (\because \gamma >1)$$, the work done is +ve i.e., work is done on gas. Also if $$\displaystyle { T }_{ f }<{ T }_{ i },w=-ve$$, thus work is done by the gas.

The adiabatic process are more steeper than isothermal process and slope of adiabatic process $$\displaystyle =\gamma \times $$slope of isothermal process and slope of adiabatic process > slope of isothermal process (since $$\displaystyle \gamma >1$$). Also the adiabatic process obey $$\displaystyle { PV }^{ \gamma }$$= constant, whereas in isothermal process PV = constant.

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The rise in temperature of an ideal gas ($$\displaystyle Y ={ 5 }/{ 3 }$$) when at $$\displaystyle { 27 }^{ o }C$$ it is adiabatically compressed to $$\dfrac{8}{27}$$ of its original volume is:

A
$$\displaystyle { 450 }^{ o }C$$

B
$$\displaystyle { 375 }^{ o }C$$

C
$$\displaystyle { 225 }^{ o }C$$

D
$$\displaystyle { 402 }^{ o }C$$

Solution

$$\displaystyle { TV }^{ \gamma -1 }=$$constant
$$\displaystyle { T }_{ 1 }[\frac { 8V }{ 27 } { ] }^{ \gamma -1 }=$$constant
$$\displaystyle \therefore \quad \frac { { T }_{ 1 } }{ T } \times [\frac { 8 }{ 27 } { ] }^{ 5/3 -1 }=1$$
or $$\displaystyle { T }_{ 1 }=[\frac { 27 }{ 8 } { ] }^{ { 2 }/{ 3 } }\times T$$
$$\displaystyle =2.25\times 300$$
$$\displaystyle =675K={ 402 }^{ o }C$$
Question 4
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A monoatomic ideal gas, initially at temperature $$T_1$$, is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature $$T_2$$ by releasing the piston suddenly. If $$L_1$$ and $$L_2$$ are lengths of gas column before and after expansion respectively, then $$\displaystyle \dfrac{T_1}{T_2}$$ is given by

A
$$\left (\displaystyle \dfrac{L_1}{L_2}\right )^{2/3}$$

B
$$\displaystyle \dfrac{L_1}{L_2}$$

C
$$\displaystyle \dfrac{L_2}{L_1}$$

D
$$\left (\displaystyle \dfrac{L_2}{L_1}\right )^{2/3}$$

Solution

In an adiabatic process $$T_1V_1 ^{\gamma -1} = T_2V_2 ^{\gamma -1}$$
For an ideal mono atomic gas the no of degrees of freedom is 3.

$$\gamma = 1 + \dfrac{2}{f}= 1 +\dfrac{2}{3}$$

$$\dfrac{T_1}{T_2} =( \dfrac{V_2}{V_1})^{\gamma-1} = ( \dfrac{V_2}{V_1})^{\dfrac{2}{3}}= ( \dfrac{L_2}{L_1})^{\dfrac{2}{3}}$$ since volume is proportional to length due to area being constant.
Question 5
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Directions For Questions

The work done during adiabatic reversible process is given by $$\displaystyle w=\dfrac { nR[{ T }_{ f }-{ T }_{ i }] }{ \gamma -1 } $$. Thus, if $$\displaystyle { T }_{ f }>{ T }_{ i }$$, then since $$\displaystyle \gamma-1=+ve\ (\because \gamma >1)$$, the work done is +ve i.e., work is done on gas. Also if $$\displaystyle { T }_{ f }<{ T }_{ i },w=-ve$$, thus work is done by the gas.

The adiabatic process are more steeper than isothermal process and slope of adiabatic process $$\displaystyle =\gamma \times $$slope of isothermal process and slope of adiabatic process > slope of isothermal process (since $$\displaystyle \gamma >1$$). Also the adiabatic process obey $$\displaystyle { PV }^{ \gamma }$$= constant, whereas in isothermal process PV = constant.

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At $$\displaystyle 27^{ o }C$$ a gas ($$\displaystyle \gamma ={ 5 }/{ 3 }$$) is compressed suddenly so that its pressure becomes $$\dfrac{1}{8}$$ of the original pressure. Final temperature of gas would be:

A
$$420K$$

B
$$300K$$

C
$$\displaystyle { -142 }^{ o }C$$

D
$$\displaystyle { 327 }^{ o }C$$

Solution

$$\displaystyle { P }^{ 1-\gamma }.{ T }^{ \gamma}=$$constant
$$\displaystyle [\frac { P }{ 8 } { ] }^{ 1-\gamma }.{ T }_{ 1 }^{ \gamma }=$$constant
$$\displaystyle \therefore \quad [\frac { 1 }{ 8 } { ] }^{ 1-\gamma }.[\frac { { T }_{ 1 } }{ T } { ] }^{ \gamma }=1$$
or $$\displaystyle [\frac { { T }_{ 1 } }{ T } { ] }^{ \gamma }=(8{ ) }^{ 1-\gamma }=(8{ ) }^{ 1-5/3 }={ 8 }^{ { -2 }/{ 3 } }$$
$$\displaystyle [\frac { { T }_{ 1 } }{ T } { ] }^{ { 5 }/{ 3 } }=(8{ ) }^{ { -2 }/{ 3 } }$$
$$\displaystyle \therefore \quad { T }_{ 1 }=T\times 0.435=300\times 0.435$$
$$\displaystyle =130.58K$$
$$\displaystyle { -142.4 }^{ o }C$$
Question 6
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Directions For Questions

One mole of a monoatomic ideal gas is taken through the cycle ABCDA as shown in the figure. $$T_A\, =\, 1000\, K$$ and $$2P_A\, =\, 3P_B\, =\, 6P_C$$.
$$\left [Assume\quad \left (\displaystyle \frac{2}{3}\right )^{0.4}\, =\, 0.85\, \ and\ \, R\, =\, \displaystyle \frac{25}{3}\, JK^{-1}\, mol^{-1}\right ]$$

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Heat lost by the gas in the process B $$\rightarrow$$ C is

A
$$5312.5 J$$

B
$$1875 J$$

C
$$0$$

D
$$8854 J$$

Solution

Given : $$T_A= 1000 K \gamma= \dfrac{5}{3} , C_v=\dfrac{3}{2}R$$ ( monoatomic gas)
Using, $$P^{1-\gamma}T^{\gamma}=constant$$ for A and B.

$$\dfrac{T_B}{T_A}= \bigg[\dfrac{P_A}{P_B}\bigg]^{\dfrac{1-\gamma}{\gamma}}$$

$$\dfrac{T_B}{T_A}= \bigg[\dfrac{3}{2}\bigg]^{-0.4}=0.85$$

$$\implies T_B=850 K$$

As for process B to C, $$V$$ is constant.
Thus $$\dfrac{T_C}{T_B}=\dfrac{P_C}{P_B}=\dfrac{1}{2}$$

$$\implies T_C=\dfrac{T_B}{2}= \dfrac{850}{2} K$$

$$Q_{BC}= \Delta U_{BC}=C_v (T_C-T_B)$$ $$(W_{BC}= 0)$$

$$Q_{BC}= \dfrac{3}{2}R\times(\dfrac{850}{2}-850) $$ $$(R=\dfrac{25}{3})$$

$$\implies Q_{BC}= -5312.5 J$$ i.e, heat is lost.
Question 7
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Directions For Questions

One mole of a monoatomic ideal gas is taken through the cycle ABCDA as shown in the figure. $$T_A\, =\, 1000\, K$$ and $$2P_A\, =\, 3P_B\, =\, 6P_C$$.
$$\left [Assume\quad \left (\displaystyle \frac{2}{3}\right )^{0.4}\, =\, 0.85\, \ and\ \, R\, =\, \displaystyle \frac{25}{3}\, JK^{-1}\, mol^{-1}\right ]$$

...view full instructions

The temperature at B is

A
$$350 K$$

B
$$1175 K$$

C
$$850 K$$

D
$$577 K$$

Solution

HINT: For an adiabatic process, $$PV^{\gamma} =$$ constant
For ideal gas, $$PV = nRT$$

$$\text{STEP 1: Given Conditions}$$
The given conditions for the thermodynamics cycle is,

$$T_A = 1000\;K$$

$$P_B = \dfrac 23 P_A$$

$$P_C = \dfrac 26 P_A = \dfrac 13 P_A$$

$$\textbf{STEP 2: Relation for Adiabatic Process}$$
The process $$A\rightarrow B$$ is an adiabatic process

=> $$PV^{\gamma} = constant \rightarrow(1)$$

Considering ideal gas equation $$PV = nRT$$

=> $$V = \dfrac{nRT}P \rightarrow(2)$$

Putting the value of $$V$$ in (1), we get

=> $$P\Big(\dfrac{nRT}P\Big)^{\gamma}=constant$$

=> $$\dfrac{P (nR)^{\gamma}T^{\gamma}}{P^{\gamma}} = constant$$

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

=> $${P_A}^{1-\gamma}{T_A}^{\gamma} = {P_B}^{1-\gamma}{T_B}^{\gamma}$$

=> $$\Big(\dfrac{P_A}{P_B}\Big)^{1-\gamma} = \Big(\dfrac{T_B}{T_A}\Big)^{\gamma}$$

=> $$\Big(\dfrac{P_B}{P_A}\Big)^{\gamma - 1} = \Big(\dfrac{T_B}{T_A}\Big)^{\gamma}$$

=> $$\Big(\dfrac{P_B}{P_A}\Big)^{1-1/\gamma} = \dfrac{T_B}{T_A}\rightarrow (3)$$

$$\textbf{STEP 3:Temperature at point B}$$
For monoatomic gas, $$\gamma = 5/3$$

Putting the value in eq (3), we get

=> $$\Big(\dfrac23\Big)^{1-\dfrac 1{5/3}}=\dfrac {T_B}{1000}$$

=> $$\Big(\dfrac 23\Big)^{1-\dfrac 35} = \Big(\dfrac{T_B}{1000}\Big)$$

=> $$\Big(\dfrac 23\Big)^{\dfrac 25} = \Big(\dfrac{T_B}{1000}\Big)$$

=> $$\Big(\dfrac 23\Big)^{0.4} = \Big(\dfrac{T_B}{1000}\Big)$$

=> $$0.85 = \Big(\dfrac{T_B}{1000}\Big)$$

=> $$0.85\times 1000 = T_B$$

=> $$T_B = 850\;K$$

Thus, option C is correct.
Question 8
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Directions For Questions

The work done during adiabatic reversible process is given by $$\displaystyle w=\dfrac { nR[{ T }_{ f }-{ T }_{ i }] }{ \gamma -1 } $$. Thus, if $$\displaystyle { T }_{ f }>{ T }_{ i }$$, then since $$\displaystyle \gamma-1=+ve\ (\because \gamma >1)$$, the work done is +ve i.e., work is done on gas. Also if $$\displaystyle { T }_{ f }<{ T }_{ i },w=-ve$$, thus work is done by the gas.

The adiabatic process are more steeper than isothermal process and slope of adiabatic process $$\displaystyle =\gamma \times $$slope of isothermal process and slope of adiabatic process > slope of isothermal process (since $$\displaystyle \gamma >1$$). Also the adiabatic process obey $$\displaystyle { PV }^{ \gamma }$$= constant, whereas in isothermal process PV = constant.

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In a particular experiment, a gas undergoes adiabatic expansion satisfying the equation $$\displaystyle { VT }^{ 3 }$$=constant. The ration of specific heats, $$\displaystyle Y $$ is:

A
$$4$$

B
$$3$$

C
$$\displaystyle \dfrac { 5 }{ 3 } $$

D
$$\displaystyle \dfrac { 4 }{ 3 } $$

Solution

$$\displaystyle \because \quad { VT }^{ 3 }=constant$$
$$\displaystyle or\quad { V }^{ { 1 }/{ 3 } }.T=constant$$
$$\displaystyle Also\quad { V }^{ { \gamma -1 } }.T=constant$$
$$\displaystyle \therefore \gamma -1=\dfrac { 1 }{ 3 } $$
$$\displaystyle \gamma =\dfrac { 4 }{ 3 } $$
Question 9
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In the given figure, let $$\Delta W_1$$ and $$\Delta W_2$$ be the work done by the system on the gas in process A and B respectively then (change in volumes in both processes is same)

A
$$\Delta W_1 > \Delta W_2$$

B
$$\Delta W_1 = \Delta W_2$$

C
$$\Delta W_1 < \Delta W_2$$

D
nothing can be said about the relation between $$\Delta W_1$$ and $$\Delta W_2$$

Solution

$$The\quad processes\quad shown\quad in\quad the\quad graph\quad are\quad both\quad isobaric\quad \\ A\quad being\quad at\quad { P }_{ 1 }\\ B\quad being\quad at\quad { P }_{ 2 }\\ Where\quad { P }_{ 2 }>{ P }_{ 1 }\\ Work\quad done\quad in\quad isobaric\quad process\quad =\quad P({ V }_{ 2 }-{ V }_{ 1 })\\ Work\quad done\quad in\quad A\quad =\quad { P }_{ 1 }({ V }_{ 2 }-{ V }_{ 1 })\\ Work\quad done\quad in\quad B\quad =\quad { P }_{ 2 }({ V }_{ 2 }-{ V }_{ 1 })\\ It\quad is\quad told\quad in\quad the\quad question\quad that\quad { V }_{ 2 }-{ V }_{ 1 }\quad is\quad the\quad change\\ in\quad volume\quad is\quad equal\quad for\quad both\quad A\quad \xi \quad B.\\ \therefore \quad \triangle { W }_{ A }\quad =\quad { P }_{ 1 }\triangle V\\ \quad \quad \quad \triangle { W }_{ B }\quad =\quad { P }_{ 2 }\triangle V\\ Dividing\quad the\quad two\quad we\quad get\quad \\ \frac { \triangle { W }_{ A } }{ \triangle { W }_{ B } } \quad =\quad \frac { { P }_{ 1 } }{ { P }_{ 2 } } \\ as\quad \quad \quad { P }_{ 2 }>{ P }_{ 1 }\\ \quad \quad \quad \quad \frac { { P }_{ 1 } }{ { P }_{ 2 } } >1\\ \quad \quad \quad \quad \frac { { P }_{ 1 } }{ { P }_{ 2 } } <1\\ \therefore \quad \frac { \triangle { W }_{ A } }{ \triangle { W }_{ B } } <1\\ \quad \quad \quad \triangle { W }_{ A }<{ W }_{ B }\\ (c)\quad \quad \triangle { W }_{ 1 }<\triangle { W }_{ 2 }$$
Question 10
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Three moles of an ideal monoatomic gas perform a cycle shown in figure. The gas temperature in different states are $$T_1=400 K, T_2=800K, T_3=2400 K,$$ and $$T_4=1200 K$$. The work done by the gas during the cycle is: $$\left [R=\dfrac {25}{3}J/mol-k\right ]$$

A
5kJ

B
10kJ

C
15kJ

D
20kJ

Solution

$$\textbf{Given}:$$ $$T_1 = 400K, T_2 = 800K, T_3 = 2400K, T_4 = 1200K$$

$$\textbf{Solution}:$$
$$W_{BA} = W_{CD} = 0 (V= const)$$
$$W_{BC}=3R T_c-T_B$$ P = const
$$W_{DA}=3R,T_A-T_D$$ P=const
$$W=3R(T_A+T_c-T_B-T_D)$$=2400R = 20KJ

$$\textbf{Hence D is the correct option}$$

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Thermodynamics Test - 75

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Thermodynamics Test - 75

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SHARING IS CARING

Question 1

$$0.036\ m^3$$

$$0.063\ m^3$$

$$0.072\ m^3$$

none

Solution

Question 2

$$\displaystyle{\dfrac{3P_0}{2^{0.4}}}$$

$$\displaystyle{\dfrac{3P_0}{2^{1.4}}}$$

$$3P_0\times 2^{0.4}$$

$$3P_0\times 2^{1.4}$$

Solution

Question 3

$$\displaystyle { 450 }^{ o }C$$

$$\displaystyle { 375 }^{ o }C$$

$$\displaystyle { 225 }^{ o }C$$

$$\displaystyle { 402 }^{ o }C$$

Solution

Question 4

$$\left (\displaystyle \dfrac{L_1}{L_2}\right )^{2/3}$$

$$\displaystyle \dfrac{L_1}{L_2}$$

$$\displaystyle \dfrac{L_2}{L_1}$$

$$\left (\displaystyle \dfrac{L_2}{L_1}\right )^{2/3}$$

Solution

Question 5

$$420K$$

$$300K$$

$$\displaystyle { -142 }^{ o }C$$

$$\displaystyle { 327 }^{ o }C$$

Solution

Question 6

$$5312.5 J$$

$$1875 J$$

$$0$$

$$8854 J$$

Solution

Question 7

$$350 K$$

$$1175 K$$

$$850 K$$

$$577 K$$

Solution

Question 8

$$4$$

$$3$$

$$\displaystyle \dfrac { 5 }{ 3 } $$

$$\displaystyle \dfrac { 4 }{ 3 } $$

Solution

Question 9

$$\Delta W_1 > \Delta W_2$$

$$\Delta W_1 = \Delta W_2$$

$$\Delta W_1 < \Delta W_2$$

nothing can be said about the relation between $$\Delta W_1$$ and $$\Delta W_2$$

Solution

Question 10

5kJ

10kJ

15kJ

20kJ

Solution

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