A cylindrical tube AB of length , closed at both ends, contains an ideal gas of 1 mol having molecular weight M. [2026]

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Solution

Q.
A cylindrical tube AB of length $l$ , closed at both ends, contains an ideal gas of 1 mol having molecular weight M. The tube is rotated in a horizontal plane with constant angular velocity $ω$ about an axis perpendicular to AB and passing through the edge at end A, as shown in the figure. If $P_{A}$ and $P_{B}$ are the pressures at A and B respectively, then

(consider the temperature is same at all points in the tube) [2026]

1 $P_{B} = P_{A} \exp (\frac{M ω^{2} l^{2}}{2 R T})$

2 $P_{B} = P_{A} \exp (\frac{M ω^{2} l^{2}}{3 R T})$

3 $P_{B} = P_{A} \exp (\frac{M ω^{2} l^{2}}{R T})$

4 $P_{B} = P_{A}$

Ans.
(1)

$A [(P + d P) - P] = (d m) (ω^{2} x)$

$d P = \frac{(d m)}{A} ω^{2} x$

$d P = \frac{(ρ) (A) (d x) ω^{2} x}{A}$

$also [P M = ρ R T]$

$ρ = \frac{P M}{R T}$

$d P = (\frac{P M}{R T}) ω^{2} x d x$

$\int_{P_{A}}^{P_{B}} \frac{d P}{P} = \frac{ω^{2} M}{R T} \int_{0}^{ℓ} x d x$

$ℓ n (\frac{P_{B}}{P_{A}}) = \frac{ω^{2} ℓ^{2} M}{2 R T}$

$P_{B} = P_{A} e^{\frac{M ω^{2} ℓ^{2}}{2 R T}}$

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