# Relation between volume and pressure of one mole of an ideal gas is given as $V^{2}=\left(\frac{P_{0}-P}{a}\right)$,

153 views
in Physics
recategorized

Relation between volume and pressure of one mole of an ideal gas is given as $V^{2}=\left(\frac{P_{0}-P}{a}\right)$, where $P_{0}$ and $a$ are positive constants. If maximum temperature attained by the gas during this process is $\frac{A P_{0}}{B R} \sqrt{\frac{P_{0}}{B a}}$, then value of $(A+B)$ is equal to . $A$ and $B$ are coprime integers) ( $R$ is universal gas constant)

by (40.9k points)

Given v2 $\left(\cfrac{p_0-p}a\right)$

p0 - p = av2

p = p- av2 ......(i)

$\because$ pv = nRT

p = $\cfrac{nRT}v$

$\cfrac{nRT}v$ = p0 - av2

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

$\cfrac{3av^2}{nR}$ = $\cfrac{p_0}{nR}$

v2$\cfrac{p_0}{3a}$

v = $\sqrt{\cfrac{p_0}{3a}}$

there value put in equation (ii)

Tmax  = $\cfrac{p_0v}{nR}$ = $\cfrac{av^3}{nR}$

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

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