Question

One mole of an ideal gas undergoes a process in which $T=T_0+a{V}^3$, where $T_0$ and 'a' are positive constants and V is volume.
minimum pressure attainable is

• A. $\frac{3}{4}\left(a^{5/3}{R}^{2/3}{T}_0^{2/3}\right)2^{1/3}$
• B. $\frac{3}{2}\left(a^{5/3}R{T}_0^{2/3}\right)3^{1/2}$
• C. $\frac{3}{2}\left(a^{1/2}{R}^{2/3}{T}_0^{3/4}\right)4^{1/3}$
• D. $\frac{3}{2}\left(a^{1/3}R{T}_0^{2/3}\right)2^{1/3}$

Answer 1

The correct option is D $\frac{3}{2}\left(a^{1/3}R{T}_0^{2/3}\right)2^{1/3}$

Given,

$n=1$

$T=T_0+a{V}^3$. . . . .(1)

From ideal gas equation,

$PV=nRT$

$T=\frac{PV}{R}$. . . . .(2)

Equating equation (1) and (2), we get,

$\frac{PV}{R}=T_0+a{V}^3$

$P=R(\frac{T_0}{V}+a{V}^2)$. . . . .(3)

For the minimum pressure $P$

$\frac{dP}{dV}=0$

$T_0(-\frac{1}{V_m^2})+2a{V}_m=0$

$V_m=(\frac{T_0}{2a}{)}^{\frac{1}{3}}$

Substituting the value of $V_m$ in equation (3), we get

$P=R\left[\frac{T_0+a{V}_m^3}{V_m}\right]$

$P=\frac{3}{2}\left[a^{1/3}R{T}_0^{2/3}\right]2^{1/3}$

The correct option is D.