Question

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gas expands its temperature according to the relation
V=KT^2/3
where K is constant. the work done by the gas when temperature changes by 60K is..

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Answer 1

$\huge\mathfrak{Answer}$

A gas expands with temperature according to relation V=KT^2/3. What is the work done when the temp changes by 30?

We know that the formula for the work done in polytropic process P*V^(n) = K’ where K is constant and n is polytropic index is given by (P1*V1 - P2*V2)/(n - 1). Now P*V^(n) can be expressed in terms of T and V as T*V^(n-1) = K’’ (where K’’=K’/n*R) by substituting P from ideal gas equation PV = nRT. Now reducing the relation given in question in the above form we get T*V^(-3/2) = K. Comparing above two equation we can obtain the value of n as -(1/2). Now the work done in the polytropic process can also be given as n*R*( T1 - T2 )/(n - 1). We know the value of R, temperature difference, n and assuming for one mole of gas we can substitute the values in the equation and solve for the work.

W = (1*8.314*30) / ( -3 / 2) = - 166.28 kJ (Assuming temperature is decreased by 30)

The work done done is -166.28 kJ when temperature decreases by 30.

Answer 2

Hi Tina here is your answer

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dW = PdV

= RT/V * dV

$= rt \div kt {}^{2 \div 3} dv$

$v = kt {}^{2 \div 3}$

Taking the derivation

$\: \: dV = K (2/3)t {}^{ - 1 \div 3}dt$

= 2/3 R (T2 - T1)

=2/3 R * 60

= 40R

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Hope it helps you..........!!

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