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Thermal Physics

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18 Mar 2010 12:48:49 IST

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Thermodynamics

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Four moles of an ideal gas obeys the law p=aV3 and its indicator diagram is shown below. The shaded area is (where a is constant)

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vijay kharya

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18 Mar 2010 15:06:22 IST

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Four moles of an ideal gas obeys the law p=aV³ and its indicator diagram is shown below. The shaded area is (where a is constant)

your question was the one given above if we find the work done (integral ( pdV) then the shaded area comes out to be 20a(V^4), but i think that the indicator diagram given is wrong because in the diagram the relation p=aV^3 does not hold true. check your question again.

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Blazing goIITian

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18 Mar 2010 15:21:15 IST

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w = 20aV ( nought)^4 ...............( work done in a polytropic process)......this is how we do it ,.......from a polytropic process eqn...we find the value of n which comes to n=-3...........now putting ths in the eqn of work done in a polytropic process..............and potting V1=V ( NOUGHT ) AND V2= 3V( NOUGHT).................i think now u can solve it!!!!!!!!!!!!!!!

vijay kharya

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18 Mar 2010 15:54:26 IST

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work done= area under the P-v curve

dW= int ( Pdv)

dw = int (aV^3 dV)

a {(V2)^4 - (V1)^4 } / 4

V2 is the final volume

and V1 is the initial volume .

W = a*{ ( 81 (v1)^4 - (v)^4) }/ 4

W = 20a*(v1)^4

blue gene plz explain your answer!!!!!!!

Nehal Wani

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21 Mar 2010 22:22:45 IST

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Guys, The answer is 3.5P_oV_o

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Blazing goIITian

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21 Mar 2010 23:01:24 IST

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20a V( nought)^4 can be written as 20 a V( nought ).P( nought)........................but wts da value of a???????????

charanjit ghai

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21 Mar 2010 23:24:05 IST

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Thermodynamics

edison

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21 Jun 2010 11:08:37 IST

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The shaded area is the work done by the ideal gas

that is W = ∫p .dV ...(1)

Here P = a V³

when P = p₀, V = V₀

Thus p₀ = a V₀³

or a = p₀ / V₀³

or P = p₀ V³/ V₀³....(2)

using (2) in (1) we obtain

W = ∫p .dV

or W =_Vo ∫ ^3Vo p₀ V³/ V₀³.dV

Now solve to obtain the area of the shaded region.

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