# NEET Questions Solved

NEET - 2014

A thermodynamic system undergoes cyclic process ABCDA as shown in figure. The work done by the system in the cycle is (a) ρoV

(b) 2ρoVo

(c) ρoVo/2

(d)zero

Concept Videos :-

#29 | Work done in Cyclic Process
#30 | Solved Example: 13

Concept Questions :-

Cyclic process

Work done in the cyclic process = Area bound by the closed configuration =Area of closed configuration

={2 [1/2 (Vo/2) x ρo]} + {-2[1/2(Vo/2)ρo]}

=zero

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NEET - 2008

If Q, E and W denote respectively the heat added, change in internal energy and the work done in a closed cyclic process, then

(a) W=0                                  (b) Q=W=0

(c) E=0                                   (d) Q=0

Concept Videos :-

#29 | Work done in Cyclic Process
#30 | Solved Example: 13

Concept Questions :-

Cyclic process

In a cyclic process, system is not isolated from the surroundings.

In a cyclic process, a system starts from one point and ends at the same point. In this case, the change in the internal energy must again be zero and therefore the thermal energy added to the system must equal the work done during the cycle. That is, in a cyclic process.

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In a cyclic process, the internal energy of the gas -

(1) Increases

(2) Decreases

(3) Remains constant

(4) Becomes zero

Concept Videos :-

#29 | Work done in Cyclic Process
#30 | Solved Example: 13

Concept Questions :-

Cyclic process

(3) Internal energy is a state function.

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A thermodynamic system is taken through the cycle PQRSP process. The net work done by the system is - (1) 20 J

(2) – 20 J

(3) 400 J

(4) – 374 J

Concept Videos :-

#29 | Work done in Cyclic Process
#30 | Solved Example: 13

Concept Questions :-

Cyclic process

(2) Work done by the system = Area of shaded portion on P-V diagram

$=\left(300-100\right){10}^{-6}×\left(200-100\right)×{10}^{3}=20\text{\hspace{0.17em}}J$

Since sense is anti-clockwise, work done is negative

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An ideal gas is taken around ABCA as shown in the above P-V diagram. The work done during a cycle is (1) 2PV

(2) PV

(3) 1/2PV

(4) Zero

Concept Videos :-

#29 | Work done in Cyclic Process
#30 | Solved Example: 13

Concept Questions :-

Cyclic process

(1) Work done = Area enclosed by triangle ABC

$=\frac{1}{2}AC×BC=\frac{1}{2}×\left(3V-V\right)×\left(3P-P\right)=2\text{\hspace{0.17em}}PV$

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An ideal gas is taken through the cycle ABCA, as shown in the figure. If the net heat supplied to the gas in the cycle is 5 J, the work done by the gas in the process CA is (1) – 5 J

(2) – 10 J

(3) – 15 J

(4) – 20 J

Concept Videos :-

#29 | Work done in Cyclic Process
#30 | Solved Example: 13

Concept Questions :-

Cyclic process

(1) For cyclic process. Total work done $={W}_{AB}+{W}_{BC}+{W}_{CA}$

ΔWAB = PΔV = 10(2 – 1) = 10J and ΔWBC =0

(as V = constant)

ΔQ = ΔU + ΔW

ΔU = 0 (Process ABCA is cyclic)

⇒ ΔQ = ΔWAB + ΔWBC + ΔWCA

⇒ 5 = 10 + 0 + ΔWCA ⇒ ΔWCA = – 5 J

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An ideal gas is taken around the cycle ABCA as shown in the P-V diagram. The net work done by the gas during the cycle is equal to (1) 12 P1V1

(2) 6 P1V1

(3) 3 P1V1

(4) 2 P1V1

Concept Videos :-

#29 | Work done in Cyclic Process
#30 | Solved Example: 13

Concept Questions :-

Cyclic process

(4) Work done $=\frac{1}{2}×2{P}_{1}×2{V}_{1}=2{P}_{1}{V}_{1}$

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Heat energy absorbed by a system in going through a cyclic process shown in figure is (1) 107 π J

(2) 104 π J

(3) 102π J

(4) 10–3π J

Concept Videos :-

#29 | Work done in Cyclic Process
#30 | Solved Example: 13

Concept Questions :-

Cyclic process

(3) In a cyclic, ΔU = 0

Δ Q = ΔU + ΔW = 0 + ΔW = Area of closed curve

$=100\text{\hspace{0.17em}}\pi ×{10}^{3}×{10}^{-3}J=100\text{\hspace{0.17em}}\pi J$

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In the diagrams (i) to (iv) , variation of volume with changing pressure is shown. A gas is taken along the path ABCD. The change in internal energy of the gas will be (1) Positive in all cases (i) to (iv)

(2) Positive in cases (i), (ii) and (iii) but zero in (iv) case

(3) Negative in cases (i), (ii) and (iii) but zero in (iv) case

(4) Zero in all four cases

Concept Videos :-

#29 | Work done in Cyclic Process
#30 | Solved Example: 13

Concept Questions :-

Cyclic process

(4) In all given cases, process is cyclic and in cyclic process ΔU = 0.

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Carbon monoxide is carried around a closed cycle abc in which bc is an isothermal process as shown in the figure. The gas absorbs 7000 J of heat as its temperature increases from 300 K to 1000 K in going from a to b. The quantity of heat rejected by the gas during the process ca is - (1) 4200 J

(2) 5000 J

(3) 9000 J

(4) 9800 J

Concept Videos :-

#29 | Work done in Cyclic Process
#30 | Solved Example: 13

Concept Questions :-

Cyclic process

(4) For path ab : ${\left(\Delta U\right)}_{ab}=7000\text{\hspace{0.17em}}J$

By using $\Delta U=\mu {C}_{V}\Delta T$

$7000=\mu ×\frac{5}{2}R×700⇒\mu =0.48$

For path ca :

${\left(\Delta Q\right)}_{ca}={\left(\Delta U\right)}_{ca}+{\left(\Delta W\right)}_{ca}$ ….(i)

${\left(\Delta U\right)}_{ab}+{\left(\Delta U\right)}_{bc}+{\left(\Delta U\right)}_{ca}=0$

$7000+0+{\left(\Delta U\right)}_{ca}=0⇒{\left(\Delta U\right)}_{ca}=-7000\text{\hspace{0.17em}}J$ ….(ii)

Also ${\left(\Delta W\right)}_{ca}={P}_{1}\left({V}_{1}-{V}_{2}\right)=\mu R\left({T}_{1}-{T}_{2}\right)$

$=0.48×8.31×\left(300-1000\right)=-2792.16\text{\hspace{0.17em}}J$ ….(iii)

on solving equations (i), (ii) and (iii)

${\left(\Delta Q\right)}_{ca}=-7000-2792.16=-9792.16\text{\hspace{0.17em}}J$ = –9800 J

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