**Given** v^{2 }= \(\left(\cfrac{p_0-p}a\right)\)

p_{0} - p = av^{2}

p = p_{0 }- av^{2} ......(i)

\(\because\) pv = nRT

p = \(\cfrac{nRT}v\)

\(\cfrac{nRT}v\) = p_{0} - av^{2}

T = \(\cfrac{p_0v}{nR}\) - \(\cfrac{av^3}{nR}\)....(ii)

\(\left(\cfrac{dT}{dv}\right)\) = 0

\(\left(\cfrac{dT}{dv}\right)\) = \(\cfrac{p_0}{nR}\) - \(\cfrac{3av^2}{nR}\) = 0

\(\cfrac{3av^2}{nR}\) = \(\cfrac{p_0}{nR}\)

v^{2} = \(\cfrac{p_0}{3a}\)

v = \(\sqrt{\cfrac{p_0}{3a}}\)

there value put in equation (ii)

T_{max } = \(\cfrac{p_0v}{nR}\) = \(\cfrac{av^3}{nR}\)

= \(\cfrac{p_0}{nR}\) \(\left(\sqrt{\cfrac{p_0}{3a}}\right)\) - \(\cfrac{a}{nR}\) \(\left({\cfrac{p_0}{3a}}\right)^{3/2}\)

T = \(\sqrt{\cfrac{p_0}{3a}}\) \(\left(\cfrac{p_0}{nR}-\cfrac{a}{nR}\times\cfrac{p_0}{3a}\right)\)

T = \(\sqrt{\cfrac{p_0}{3a}}\) \(\left(\cfrac{3p_0-p_0}{3nR}\right)\)

T = \(\cfrac{2p_0}{3nR}\) \(\sqrt{\cfrac{p_0}{3a}}\)

Give T = \(\cfrac{A\,p_0}{B\,nR}\) \(\sqrt{\cfrac{p_0}{Ba}}\)

then A = 2

B = 3

Then value at A + B equal to

= 2 + 3

= 5