Thermodynamics Test

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Thermodynamics Test - 76

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Weekly Quiz Competition

Question 1
1 / -0

When an ideal gas undergoes an adiabatic change causing a temperature change of $$\Delta$$T, then which of the following statements are correct
(i) there is no heat gained or lost by the gas
(ii) the work done by the gas is equal to change in internal energy.
(iii) the change in internal energy per mole of the gas is $$C_v \times \Delta T$$, where $$C_v$$ is the molar heat capacity at constant volume

A
(i), (ii), (iii)

B
(i), (ii)

C
(i), (iii)

D
(i)

Solution

In adiabatic process Q = 0
$$\therefore$$ No heat is gained or lost,
$$\triangle v = Q-W$$
$$Q = 0$$
$$\therefore \triangle V = W$$
Change in internal energy cannot be calculated.
Directly, we used to use the first law of thermadyherin, which here given.
$$\triangle v = W$$
ANSWER : (B).
Question 2
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One mole of a gas is subjected to two process AB and BC, one after the other as shown in the figure. BC is represented by $$PV^n = Constant$$. We can conclude that $$(T = temperature, W = work$$ done by gas, $$V = volume$$ and $$U = internal \ energy)$$

A
$$T_A=T_B=T_C$$

B
$$V_A < V_B, P_B < P_C$$

C
$$W_{AB} > W_{BC}$$

D
$$U_A < U_B$$

Solution

$$\textbf{Solution}:$$
By PV = nRT
$$T_B > T_A$$
$$W_{AB} = P_0(2V_0-V_0)= P_0 V_0$$
$$W_{BC}= RT1n\dfrac{3}{2} = 2P_0 V_0[1n3-1n2] =2P_0 V_0(2.3030)(0.477-0.30) =0.8142 P_0V_0 < W_{AB}$$

$$\textbf{Hence C is the correct option}$$
Question 3
1 / -0

Two samples of gases 1 and 2 are initially kept in the same state. Sample 1 is expanded through an isothermal process whereas sample 2 through an adiabatic process up to the same final volume. The final temperature in process 1 and 2 are $$T_1$$ and $$T_2$$ respectively, then

A
$$T_1 > T_2$$

B
$$T_1 = T_2$$

C
$$T_1 < T_2$$

D
the relation between $$T_1$$ and $$T_2$$ cannot be deduced

Solution

The work done is more in isothermal process than adiabatic process.
Let the equal heat given to both be Q.
Work done by 1 is $${ W }_{ 1 }$$.
Work done by 2 is$$ { W }_{ 2 }$$.
where $${ W }_{ 1 }>{ W }_{ 2 }$$
Let the initial temperature be T,
For 1 : the $${ T }_{ 1 }$$ will be same as T because it is an isothermal
process.
$$\therefore \triangle v = 0$$
For 2 : $$P{ V }^{ 4 } = K$$
$$T{ V }^{ 4-1 } = K$$
$${ T }_{ 1 }{ V }_{ 1 }^{ 4-1 } = { T }_{ 2 }{ V }_{ 2 }^{ 4-1 }$$
$$\dfrac { { T }_{ 1 } }{ { T }_{ 2 } } ={ (\dfrac { { V }_{ 2 } }{ { V }_{ 1 } } ) }^{ 4-1 }$$
As the gas expands : $${ V }_{ 2 }>{ V }_{ 1 } ; \dfrac { { V }_{ 2 } }{ { V }_{ 1 } } >1$$
$$\dfrac { { T }_{ 1 } }{ { T }_{ 2 } } >1,$$
$$\therefore { T }_{ 1 }>{ T }_{ 2 }$$
Hence, OPTION : A.
Question 4
1 / -0

Two samples of gases 1 and 2 are initially kept in the same state. Sample 1 is expanded through an isothermal process whereas sample 2 through an adiabatic process up to the same final volume. Let $$P_1$$ and $$P_2$$ be the final pressure of the samples 1 and 2 respectively then

A
$$P_1 < P_2$$

B
$$P_1=P_2$$

C
$$P_1 > P_2$$

D
the relation between $$P_1$$ and $$P_2$$ cannot be deduced

Solution

For isothermal process temperature is constant,
$$\therefore { P }_{ O }{ V }_{ O } = { P }_{ 1 }{ V }_{ 1 }$$
$${ P }_{ 1 } = \dfrac { { P }_{ 0 }{ V }_{ 0 } }{ { V }_{ 1 } } \longrightarrow (1)$$
For adiabatic process,
$${ P }_{ 0 }{ V }_{ 0 }^{ 4 } = { P }_{ 2 }{ V }_{ 2 }^{ Y }$$
$${ P }_{ 2 } = { P }_{ 0 }{ \left[ \dfrac { { V }_{ 0 } }{ { V }_{ 2 } } \right] }^{ Y }$$
Given, $${ V }_{ 2 } = { V }_{ 1 }$$
$${ P }_{ 2 } = { P }_{ 0 }{ \left[ \dfrac { { V }_{ 0 } }{ { V }_{ 2 } } \right] }^{ Y }\longrightarrow (2)$$
dividing (1) and (2)
$$\dfrac { { P }_{ 1 } }{ { P }_{ 2 } } = \dfrac { \dfrac { { P }_{ 0 }{ V }_{ 0 } }{ { V }_{ 1 } } }{ { P }_{ 0 }{ \left[ \dfrac { { V }_{ 0 } }{ { V }_{ 1 } } \right] }^{ Y } }$$
$$\dfrac { { P }_{ 1 } }{ { P }_{ 2 } } = { \left[ \dfrac { { V }_{ 0 } }{ { V }_{ 1 } } \right] }^{ 1-Y }$$
As the gas in expanded, $${ V }_{ 0 }<{ V }_{ 1 }$$
$$\dfrac { { V }_{ 0 } }{ { V }_{ 1 } } <1$$
As, $$Y>1$$
$$1-Y<0$$
$$\therefore { \left[ \dfrac { { V }_{ 0 } }{ { V }_{ 1 } } \right] }^{ 1-Y }>1$$
$$\dfrac { { P }_{ 1 } }{ { P }_{ 2 } } > 1$$
$$\therefore { P }_{ 1 }>{ P }_{ 2 } (C)$$
OPTION : C
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.

...view full instructions

Which of the following shows adiabatic process in the figures given below?

A
II. III

B
I, III

C
II, IV

D
I, IV

Solution

Adiabatic slopes are more steeper than isothermal.

Isothermal change:
$$\displaystyle PV=K$$ (at constant $$T$$)
$$\displaystyle \partial P.V+P.\partial V=0$$
$$\displaystyle \therefore \quad slope=[\dfrac { \partial P }{ \partial V } ]=-\dfrac { P }{ V } $$

Adiabatic change:
$$\displaystyle { PV }^{ \gamma }=K$$
$$\displaystyle \partial P.{ V }^{ \gamma }+PY{ V }^{ \gamma-1 }\partial V=0$$
$$\displaystyle \therefore \quad slope=[\dfrac { \partial P }{ \partial V } ]=-\dfrac { \gamma{ V }^{ \gamma-1 }.P }{ { V }^{ \gamma } } =-u.\dfrac { P }{ V } $$
$$\displaystyle \therefore \dfrac { Adiabatic\quad slope }{ Isothermal\quad slope } =\gamma>1$$
Question 6
1 / -0

Separation between the plates of a parallel plate capacitor connected to a battery (zero resistance) of constant e.m.f. is increased with constant (very slow) speed by external forces During the process W is the work done by external forces $$\displaystyle \Delta U $$ is the change in potential energy of the capacitor, $$\displaystyle W_{b}$$ is the work done by the battery and H is the heat loss in the circuit Then

A
$$\displaystyle W+W_{b}=\Delta U$$

B
$$\displaystyle H\neq 0$$

C
$$\displaystyle H=\Delta U$$

D
W = 0

Solution

Since there will be some loss in form of electromagnetic radiations, hence energy conservation equation must take into account that loss. Option A is incorrect.
Heat loss will be 0 since no resistor is present.
Change in potential energy will be non zero since capacitance changes and hence energy by $$\dfrac { 1 }{ 2 } C{ V }^{ 2 }$$
Option D is correct.
Question 7
1 / -0

If $$\gamma $$ denotes the ratio of the two specific heats of a gas, the ratio of the slopes of adiabatic and isothermal $$p-V$$ curves at their point of intersection is

A
$${ 1 }/{ \gamma }$$

B
$$\gamma $$

C
$$\gamma -1$$

D
$$\gamma +1$$

Solution

Isothermal process : $$PV = $$ constant
Differentiating, we get $$PdV+VdP =0$$
$$\implies$$ Slope of isothermal curve $$\bigg(\dfrac{dP}{dV}\bigg)_{iso} = \dfrac{-P}{V}$$ ...........(1)

Adiabtic process : $$PV ^{\gamma}= $$ constant
Differentiating, we get $$P \gamma V^{\gamma-1}dV+V^{\gamma}dP =0$$
$$\implies$$ Slope of adiabatic curve $$\bigg(\dfrac{dP}{dV}\bigg)_{adi} = \dfrac{-\gamma P}{V}$$ ...........(2)
$$\implies$$ Ratio of slopes $$\dfrac{(\frac{dP}{dV})_{adi}}{(\frac{dP}{dV})_{iso}} = \dfrac{(-\gamma P/V)}{-P/V} =\gamma$$
Question 8
1 / -0

A reversible engine converts one-sixth of the heat supplied into work. When the temperature of the sink is reduced by $$\displaystyle { 62 }^{ \circ }C$$, the efficiency of the engine is doubled. The temperature of the source and sink are:

A
$$\displaystyle { 99 }^{ \circ }C,{ 37 }^{ \circ }C$$

B
$$\displaystyle { 80 }^{ \circ }C,{ 37 }^{ \circ }C$$

C
$$\displaystyle { 95 }^{ \circ }C,{ 37 }^{ \circ }C$$

D
$$\displaystyle { 90 }^{ \circ }C,{ 37 }^{ \circ }C$$

Solution

Given : Work $$W = \dfrac{1}{6}Q_H$$
$$\therefore$$ Efficiency of engine $$\eta = \dfrac{W}{Q_H} = \dfrac{1}{6}$$ .........(1)
Let the initial source and sink temperatures be $$T_H$$ and $$T_L$$ respectively.
$$\therefore$$ $$\eta =1 - \dfrac{T_L}{T_H}$$

or, $$\dfrac{1}{6} =1 - \dfrac{T_L}{T_H}$$ $$\implies T_H = 1.2T_L$$ ........(2)

Now the sink temperature is reduced by $$62^o C$$ i.e $$T_L' = T_L - 62$$ ........(3)
New efficiency is doubled i.e $$\eta ' =\dfrac{2}{6} = \dfrac{1}{3}$$
Using $$\eta' =1- \dfrac{T_L' }{T_H}$$

$$\therefore$$   $$\dfrac{1}{3} =1- \dfrac{T_L - 62}{1.2 T_L}$$ (using (2) and (3))

or,   $$ \dfrac{T_L - 62}{1.2 T_L} = \dfrac{2}{3}$$ $$\implies T_L = 310 K$$ or, $$T_L = 320 - 273 = 37^o C$$

$$\therefore$$ $$T_H = 1.2 \times 310 = 372 K$$
$$\implies$$ $$T_H = 372 - 273 = 99^o C$$
Question 9
1 / -0

$$200\ g$$ water is heated from $$40^oC$$ to $$60^oC$$. Ignoring the slight expansion of water, the change in its internal energy is close to:
(Given specific heat of water $$= 4184\ J/kg/K$$)

A
$$167.4\ kJ$$

B
$$16.7\ kJ$$

C
$$8.4\ kJ$$

D
$$4.2\ kJ$$

Solution

Since expansion is ignored, work done is zero and hence by first law of thermodynamics, internal energy change equals heat supplied.

$$\Delta U=Q=0.2(4184)(20)\ J=16.7\ kJ$$
Question 10
1 / -0

In a thermodynamic processes, If the amount of work done on the gas by its surrounding is 320 J and the internal energy is increased by 560 J. Calculate the how much heat is transferred between the gas and its surrounding.

A
240 J absorbed

B
240 J dissipated

C
880 J absorbed

D
880 J dissipated

E
None of the above

Solution

From first law of thermodynamic, $$Q=\Delta U+W$$
As the work done on the system so it will be negative i.e, $$W=-320 J$$ and since internal energy is increased so it would be positive i.e, $$\Delta U=560 J$$
Thus, $$dQ=560-320=240 J$$
As $$Q$$ is positive so the system will absorb heat.

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Thermodynamics Test - 76

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Thermodynamics Test - 76

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Question 1

(i), (ii), (iii)

(i), (ii)

(i), (iii)

(i)

Solution

Question 2

$$T_A=T_B=T_C$$

$$V_A < V_B, P_B < P_C$$

$$W_{AB} > W_{BC}$$

$$U_A < U_B$$

Solution

Question 3

$$T_1 > T_2$$

$$T_1 = T_2$$

$$T_1 < T_2$$

the relation between $$T_1$$ and $$T_2$$ cannot be deduced

Solution

Question 4

$$P_1 < P_2$$

$$P_1=P_2$$

$$P_1 > P_2$$

the relation between $$P_1$$ and $$P_2$$ cannot be deduced

Solution

Question 5

II. III

I, III

II, IV

I, IV

Solution

Question 6

$$\displaystyle W+W_{b}=\Delta U$$

$$\displaystyle H\neq 0$$

$$\displaystyle H=\Delta U$$

W = 0

Solution

Question 7

$${ 1 }/{ \gamma }$$

$$\gamma $$

$$\gamma -1$$

$$\gamma +1$$

Solution

Question 8

$$\displaystyle { 99 }^{ \circ }C,{ 37 }^{ \circ }C$$

$$\displaystyle { 80 }^{ \circ }C,{ 37 }^{ \circ }C$$

$$\displaystyle { 95 }^{ \circ }C,{ 37 }^{ \circ }C$$

$$\displaystyle { 90 }^{ \circ }C,{ 37 }^{ \circ }C$$

Solution

Question 9

$$167.4\ kJ$$

$$16.7\ kJ$$

$$8.4\ kJ$$

$$4.2\ kJ$$

Solution

Question 10

240 J absorbed

240 J dissipated

880 J absorbed

880 J dissipated

None of the above

Solution

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