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Related Videos :-

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For compression:-

${\mathrm{W}}_{2}>{\mathrm{W}}_{1}\mathrm{or}{\mathrm{W}}_{1}{\mathrm{W}}_{2}$

Work done = area under P - V curve.

Work done is more for adiabatic compression.

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NEET - 2016

A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half .Then -

(a) compressing the gas through adiabatic process will require more work to be done

(b) compressing the gas isothermally or adiabatically will require the same amount work

(c) which of the case (whether compression through isothermal or through adiabatic process) requires more work will depend upon the atomicity of the gas

(d) compressing the gas isothermally will require more work to be done

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(a) The solution of this question can be understood by plotting a p-V graph for the compression of a gas isothermally and adiabatically simultaneously to half of its initial volume, ie.

Since, the isothermal curve is less steeper than the adiabatic curve. So. area under the p-V curve for adiabatic process has more magnitude than isothermal curve. Hence, work done in adiabatic process will be more than in isothermal process

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NEET - 2015

An ideal gas is compressed to half its initial volume by means of several process. Which of the process results in the maximum work done on the gas?

(a) Adiabatic

(b) Isobaric

(c) Isochoric

(d) Isothermal

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Given, ideal gas is compressed to half its inital volume i.e

V_{o}=V/2

The isochoric process is one in which volume is kept constant, meaning that work done by the system will be zero. i.e. W_{isochoric}=0

As we know, work done on the gas=Area under the curve i.e.

W_{adiabatic}>W_{isothermal}>W_{isobaric}

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NEET - 2013

A gas is taken through the cycle A→B→C→A, as shown. What is the net work done by the gas?

(a)2000J

(b)1000J

(c)Zero

(d)-2000J

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(b) Net work done=Area enclosed in pV curve,i.e., ΔABC

=1/2 x 5 x 10^{-3} x 4 x 10^{5} J

=10^{3} J

=1000

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A cylinder fitted with a piston contains 0.2 *moles* of air at temperature 27°*C*. The piston is pushed so slowly that the air within the cylinder remains in thermal equilibrium with the surroundings. Find the approximate work done by the system if the final volume is twice the initial volume** **

(1) 543 *J*

(2) 345 *J*

(3) 453 *J*

(4) 600 *J*

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(2) $W=\mu RT{\mathrm{log}}_{e}\left(\frac{{\displaystyle {V}_{2}}}{{\displaystyle {V}_{1}}}\right)$

$=0.2\times 8.3\times {\mathrm{log}}_{e}2\times (27+273)$

$=0.2\times 8.3\times 300\times 0.693=345J$

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Two identical samples of a gas are allowed to expand (i) isothermally (ii) adiabatically. Work done is** **

(1) More in the isothermal process

(2) More in the adiabatic process

(3) Neither of them

(4) Equal in both processes

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(1) In thermodynamic processes.

Work done = Area covered by *PV* diagram with *V*-axis

From graph it is clear that ${\text{(Area)}}_{\mathrm{iso}}>{\left(\mathrm{Area}\right)}_{\mathrm{adi}}$

⇒ ${W}_{iso}>{W}_{adi}$

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