Question

A ring shaped tube contains two ideal gases the equal masses and molar masses,M1=32 and M2=28. The gases are separated by one fixed position and another movable stopper S which can move freely without friction inside the ring. The angle between the fixed partition and movable partition is?

Answer 1

Masses of both the given gases are equal .
Let w is the mass of each gas .
Number of mole of gas 1 = weight/M₁ = w/32
Number of mole of gas 2 = weight/M₂ = w/28

Here Let CSA of tube is A₀ . And Length of tube in which gas 1 include is L₁ and length of tube in which gas 2 is include L₂ .
So, volume of gas 1 = CSA × length = A₀L₁
volume of gas 2 = A₀L₂

Because temperature and pressure is constant
So, use formula, V₁n₂ = V₂n₁ [ As you know, PV = nRT ]
A₀L₁(w/28) = A₀L₂ (w/32)
L₁/L₂ = 28/32 = 7/8

Now, use θ = L/R concept ,
Then angle made by gas 1 = L₁/(L₁ + L₂) × 360° = 7/(7 + 8) × 360 = 168°
and angle made by gas 2 = 360 - 168 = 192°

Answer 2

Answer:

        let the cross section area is A0 and common temp T0, common pressure P0,for gas 1: PV=nRTP0 * A0 * L1 = [w/32] R * T0 ...............(1)for gas 2: P0 * A0 * L2 = [w/28] * R * T0..............(2)where L indicates the length of the arc on ring By equation 1 / 2L1/L2 = 28/32 = 7/8So angle made by gas (1) = 7/[7+8] * 360= 7/15*360 = 168 degree

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Explanation: