Question

The pressure of one mole of an ideal gas varies acorrding to the law P= (Po -- aV^2 ) where Po and a are positive constants temprature. The hihest temprature that the gas may attain is ?

Answer 1

Pressure of one mole of an ideal gas varies according to the law , $P=P_0-aV^2$

we know, PV = nRT

n = 1 , P = Po - aV²

so, (Po - aV²)V = RT

or, PoV - aV³ = RT

or, T = (Po/R)V - (a/R)V³

differentiate T , with respect to V,

dT/dV = (Po/R) - 3(a/R)V²

at , dT/dV = 0, (Po/R) = 3(a/R)V²
V² = Po/3a,
V = $\sqrt{\frac{P_0}{3a}}$

again, differentiate with respect to V,

d²T/dV² = 0 - 6(a/R)V
at V = $\sqrt{\frac{P_0}{3a}}$ , d²T/dV² < 0 so, at V = $\sqrt{\frac{P_0}{3a}}$, T attains maximum.

so, highest Temperature , T = $\frac{P_0}{R}\sqrt{\frac{P_0}{3a}}-\frac{a}{R}\left(\sqrt{\frac{P_0}{3a}}\right)^3$