Question

In a non flow reversible process for which p = (-3v + 15)*10^5 n/m^2, v changes from 1 m^3 to m^3. The work done

Answer 1

Answer:

Explanation:

Just integrate

Integral P.dV where limit goes from 1 to 3

Answer 2

The work done by a non flow reversible process is $10.5\times 10^5\ J$.

Explanation:

The non flow reversible process for which :

$P=(-3V+15)\times 10^5\ N/m^2$

V is changing from 1 m³ to 2 m³. The work done in non flow reversible process is given by :

$W=\int\limits^a_b {PdV} \\\\W=\int\limits^2_1 {((-3V+15)\times 10^5)dV} \\\\W=(\dfrac{-3V^2}{2}+15V)\times 10^5|_1^2\\\\W=(\dfrac{-3(2)^2}{2}+15(2))\times 10^5-(\dfrac{-3(1)^2}{2}+15(1))\times 10^5\\\\W=10.5\times 10^5\ J$

So, the work done by a non flow reversible process is $10.5\times 10^5\ J$.

Learn more,

Non flow reversible process

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