Question

Find the minimum attainable pressure of an ideal gas in process $T = T_{0} + \alpha V^{2}$ , where $T_{0}$ and $\alpha $ are positive constants and $V$ is the volume of one mole of gas.

• A. $2 R \sqrt{\alpha T_{0}}$
• B. $3 R \sqrt{\alpha T_{0}}$
• C. $3 R$
• D. $3 R \sqrt{\frac{\alpha T_{0}}{2}}$

Answer 1

Answer: (A)

Given,
$T = T_{0} + \alpha V^{2}$ ...(i)
For 1 mol of a gas, $P V = R T$
of $V = \frac{R T}{P}$
Substituting this value in Eq. (i), we get
$T = T_{0} + \alpha \left(\frac{R T}{P}\right)^{2} = T_{0} + \alpha \frac{R^{2} T^{2}}{P^{2}}$
or $T P^{2} = T_{0} P^{2} + \alpha R^{2} T^{2}$
or $P = \sqrt{\alpha } R T \left(T - T_{0}\right)^{- 1 / 2}$ ...(ii)
After differentiating, we get
$\frac{d P}{d T} = \sqrt{\alpha } R \left[\left(T - T_{0}\right)^{- 1 / 2} - \frac{1}{2} T \left(T - T_{0}\right)^{- 3 / 2}\right]$
For minimum pressure,
$\frac{d P}{d T} = 0$
$∴ \, 0 = \sqrt{\alpha } R \left[\left(\text{T} - \left(\text{T}\right)_{0}\right)^{- 1 / 2} - \frac{1}{2} \text{T} \left(\text{T} - \left(\text{T}\right)_{0}\right)^{- 3 / 2}\right]$
After solving, T = 2T₀
From Eq. (ii),
$P_{min} = \sqrt{\alpha } R 2 T_{0} \left(2 T_{0} - T_{0}\right)^{- 1 / 2}$
$= 2 R \sqrt{\alpha T_{0}}$