NEET - 2014

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

(a) ρ_{o}V_{o }

(b) 2ρ_{o}V_{o}

(c) ρ_{o}V_{o}/2

(d)zero

Concept Questions :-

Cyclic process

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

={2 [1/2 (V_{o}/2) x ρ_{o}]} + {-2[1/2(V_{o}/2)ρ_{o}]}

=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 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.

$\u2206U=0orE=0$

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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 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 Questions :-

Cyclic process

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

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

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) 2*PV*

(2) *PV *

(3) 1/2*PV *

(4) Zero

Concept Questions :-

Cyclic process

(1) Work done = Area enclosed by triangle *ABC*

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

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An ideal gas is taken through the cycle *A* → *B* → *C* → *A*, 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 *C* → *A* is

(1) – 5 *J*

(2) – 10 *J*

(3) – 15 *J *

(4) – 20 *J*

Concept Questions :-

Cyclic process

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

Δ*W _{AB}* =

(as *V* = constant)

Δ*Q* = Δ*U* + Δ*W*

Δ*U* = 0 (Process *ABCA* is cyclic)

⇒ Δ*Q* = Δ*W _{AB}* + Δ

⇒ 5 = 10 + 0 + Δ*W _{CA}* ⇒ Δ

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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 *P*_{1}*V*_{1}

(2) 6 *P*_{1}*V*_{1}

(3) 3 *P*_{1}*V*_{1}

(4) 2 *P*_{1}*V*_{1}

Concept Questions :-

Cyclic process

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

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Heat energy absorbed by a system in going through a cyclic process shown in figure is

(1) 10^{7} *π J*

(2) 10^{4} *π J*

(3) 10^{2}*π J *

(4) 10^{–3}*π J*

Concept Questions :-

Cyclic process

(3) In a cyclic, Δ*U* = 0

Δ *Q* = Δ*U* + Δ*W* = 0 + Δ*W* = Area of closed curve

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

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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 Questions :-

Cyclic process

(4) In all given cases, process is cyclic and in cyclic process Δ*U* = 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 t*o 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 Questions :-

Cyclic process

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

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

$7000=\mu \times \frac{5}{2}R\times 700\Rightarrow \mu =0.48$

For path *ca* :

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

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

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

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

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