Thermodynamics Test

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Thermodynamics Test - 51

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Weekly Quiz Competition

Question 1
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A sample of a diatomic gas at pressure $$p$$, volume $$V$$ and temperature $$T$$ is compressed to $$V/2$$ isothermally. Its pressure is found to be $$p_{i}$$. If it is compressed to $$V/2$$ adiabatically, its pressure comes out to be $$p_{a}$$. What is the relation between $$p_{i}$$ and $$p_{a}$$?

A
$$p_{i} = p_{a}$$

B
$$2p_{i} = p_{a}$$

C
$$p_{i} = 2.6p_{a}$$

D
$$2.6p_{i} = 2p_{a}$$

Solution

Given : A sample of a diatomic gas at pressure p, volume V and temperature T is compressed to V/2 isothermally. Its pressure is found to be pi. If it is compressed to V/2 adiabatically, its pressure comes out to be pa. What is the relation between pi and pa

Solution :
Case 1(Isothermal process)
$$V_1= V$$ , $$V_2=\frac{V}{2}$$ , $$P_1=P$$ , $$P_2=P_i$$
$$P_1V_1=P_2V_2$$
$$PV=P_i \frac{V}{2}$$
$$P_i=2P$$

Case 2(adiabatic process)
$$V_1= V$$ , $$V_2=\frac{V}{2}$$ , $$P_1=P$$ , $$P_2=Pa$$
$$\gamma =1.4$$
$$PV^{\gamma} = constant$$
$$PV^{\gamma} =(P_a \frac{V}{2})^{\gamma}$$
$$P_a =P (2)^{1.4}$$
$$P_a = 2.6P$$

The Correct Opt = D
Question 2
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A sample of diatomic gas at pressure p, volume V and temperature T is compressed to V/2 isothermally. Its pressure is found to be $${p}_{i}$$. If it is compressed to V/2 adiabaticaly, its pressure comes out to be $${p}_{a}$$. What is the relation between $${p}_{i}$$ and $${p}_{a}$$?

A
$${p}_{i}$$ = $${p}_{a}$$

B
$$2{p}_{i}=$$ $${p}_{a}$$

C
$${p}_{i}$$ = $$2\cdot 6{p}_{a}$$

D
$$2\cdot 6{p}_{i}$$ = $$2{p}_{a}$$

Solution

Given:
a diatomic gas,
Pressure = p; Volume =V; Temperature =T
Compressed isothermally
$$v_i=\dfrac V2$$ and pressure = $$p_i$$
Compressed adiabatically,
$$v_a=\dfrac V2$$ and pressure = $$p_a$$
To find:
Relation between $$p_i$$ and $$p_a$$

We know,
For isothermal,
$$\dfrac {P_1}{P_2}=\dfrac {V_2}{V_1}\\\implies \dfrac {p_i}{p}=2\\\implies p=\dfrac {p_i}2......(i)$$

For adiabatic,
$$P_1V_1^{\gamma}=P_2V_2^{\gamma}\\\implies p_a\left(\dfrac V2\right)^{\gamma}=p( V)^{\gamma}\\\implies p_a=p\left(V\times \dfrac 2V\right)^{1.4}\\\implies p_a=p(2)^{1.4}\\\implies p_a=p(2.6)\\\implies p=\dfrac {p_a}{2.6}.....(ii)$$
Equating equation (i) and (ii), we get,
$$\dfrac {p_i}{2}=\dfrac {p_a}{2.6}\\\implies 2.6p_i=2p_a$$
is the correct answer.
Question 3
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Three stars A, B, C have surface temperatures $$T_A$$, $$T_B$$ and $$T_C$$. A appears bluish, B appears reddish and C appears yellowish. We can conclude that:

A
$$T_A > T_C > T_B$$

B
$$T_A > T_B > T_C$$

C
$$T_B > T_C > T_A$$

D
$$T_C > T_B > T_A$$

Solution

According to Wien's Displacement law we know that:-
$$\lambda=\dfrac bT \ldots \ldots (1)$$.

Given that three stars $$A, B $$ and $$C$$ where $$A$$ appears bluish, $$B$$ appears reddish and $$C$$ appears yellowish.

We know that $$\lambda_{blue}< \lambda_{yellow}< \lambda_{Red}$$
$$\implies \lambda_{A}<\lambda_{C}<\lambda_{B} ..... (2)$$

from (1) and (2) we get
$$\implies T_{A} > T_{C}> T_{B}$$

Hence option $$(A)$$ is correct
Question 4
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Which of the following P-V diagram representing the graph of isomeric process?

A

B

C

D

Solution

ISomeric process is equal to isochoric process. i.e, volume is constant.
$$PV=nRT\\ \cfrac{P}{T}=\cfrac{nR}{V}\\ \cfrac{P}{T}=K$$
Question 5
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A sample of ideal gas $$ (\gamma = 1.4) $$ is heated at constant pressure. If an amount of $$100\ J$$ heat is supplied to the gas, the work done by the gas is

A
$$28.57\ J$$

B
$$56.28\ J$$

C
$$36.23 J$$

D
$$42.12 J$$

Solution

Given,
$$\gamma=1.4$$
$$\Delta Q=100J$$
For ideal gas,
Change in heat,
$$\Delta Q=nC_P(T_2-T_1) $$. . . . . . . . .(1)
Change in internal energy,
$$\Delta U=nC_V(T_2-T_1)$$. . . . . . .. . . . .(2)
Divide equation (2) by equation (1), we get
$$\dfrac{\Delta U}{\Delta Q}=\dfrac{nC_V(T_2-T_1)}{nC_P(T_2-T_1)}=\dfrac{C_V}{C_P}$$

$$\Delta U=\dfrac{\Delta Q}{\gamma}$$

$$\Delta U=\dfrac{100}{1.4}=71.42J$$
From first law of thermodynamics,
$$\Delta Q=\Delta U+\Delta W$$
$$\Delta W=\Delta Q-\Delta U$$
$$\Delta W=100-71.42$$
$$\Delta W=28.57J$$
The correct option is A.
Question 6
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The possibility of increase in the temperature of a gas without adding heat to it happens in :

A
adiabatic expansion

B
isothermal expansion

C
adiabatic compression

D
isothermal compression

Solution

It is possible to increase the temperature of a gas without adding heat to it, during adiabatic compression the temperature of a gas increases while no heat is given to it.

For an adiabatic compression, no heat is given or taken out in adiabatic process.

Therefore, $$\Delta Q = 0$$

According to the first law of thermodynamics,

$$\Delta Q=\Delta U+\Delta W$$

$$\Delta U = -\Delta W ( \Delta Q =0)$$

In compression work is done on the gas, i.e. work done is negative. Therefore,

$$\Delta U =$$ Positive

Hence, internal energy of the gas increases due to which its temperature increases.
Question 7
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Match the column I with column II
Types of processes Features
(A)Isothermal (p) $$\Delta$$ Q = 0
(B)Isobaric (q) Volume constant
(C)Isochoric (r) Pressure constant
(D)Adiabatic (s) Temperature constant

A
(A) - (s), (B) - (r), (C) (q), (D) (p)

B
(A) - (p), (B) - (s), (C) (r), (D) (q)

C
(A) - (q), (B) - (r), (C) (p), (D) (s)

D
(A) - (r), (B) - (p), (C) (q), (D) (s)

Solution

Isothermal -----Temperature is constant
Isobaric ------Pressure is constant
Isochoric----Volume is constant
Adiabatic ----$$\Delta Q=0$$ There is no heat exchange.
Question 8
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A heat insulating cylinder with a movable piston contains 5 moles of hydrogen at standard temperature and pressure if the gas is compressed to quarter of its original volume when the pressure of the gas is increased by ($$\gamma$$ = 1.4 )

A
$${ 2 }^{ 1.4 }$$

B
$${ 3 }^{ 1.4 }$$

C
$${ 4 }^{ 1.4 }$$

D
$${ 5 }^{ 1.4 }$$

Solution

Given,
$$n=5mole$$
$$V_1=V$$
$$V_2=\dfrac{V}{4}$$
$$\gamma=1.4$$
$$P_1=P$$

For adiabatic process, gas equation is
$$PV^{\gamma}=K$$ (constant)
$$P_1V_1^{\gamma}=P_2 V_2^{\gamma}$$

$$P_2=P_1\left(\dfrac{V_1}{V_2}\right)^{\gamma}$$

$$P_2=\left(\dfrac{V}{V/4}\right)^{\gamma}P$$

$$P_2=4^{1.4}P$$
The correct option is C.
Question 9
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Let $$E_D, E_P, E_S$$ denote efficiency of a diesel, a petrol and a steam engine respectively. which of the following is correct?

A
$$E_D < E_P > E_S$$

B
$$E_D < E_P < E_S$$

C
$$E_P > E_D > E_S$$

D
$$E_P < E_S < E_D$$

Solution

Let, $$E_D,E_P,E_S$$ denote efficiency of a diesel, a petrol stream engine respective.
$$E_P>E_D>E_S$$
because petrol and diesel using the Carnot cycle.
Question 10
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The most efficient engine is?

A
Petrol

B
Electric

C
Steam

D
Diesel

Solution

The electric engine is most efficient. Electric motors are very efficient at converting electricity into work. Physicists throw around abstract numbers like 90%, whereas when discussing combustion engines, they use numbers less than 40%. But you cannot compare the efficiency of converting electricity into motion to the efficiency of converting gasoline into motion.