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Question

A ring shaped tube contains two ideal gases with equal masses and molecular weight m1=32 g and m2=28 g. The gases are separated by one fixed partition P and another movable conducting partition S which can move freely without friction inside the ring. The angle α as shown in the figure in equilibrium is:

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Solution

The correct option is **D** 16π15

At the condition of equilibrium, the pressure of gas on both sides of movable partition S will be equal.

⇒P1=P2

Also the partition S is conducting hence it will maintain the thermal equilibrium of both gases by allowing exchange of heat between then.

Hence at the equilibrium position,

T1=T2

Considering the thermodynamic process to be reversible. because friction is absent,

From ideal gas equation,

P1=P2

n1RT1V1=n2RT2V2

n1V1=n2V2(∵T1=T2).....(i)

Equal mass of both the gases is present inside tube,

n1=mm1& n2=mm2

Also the area of cross - section of tube is constant.

⇒ volume of the gas =A×l

(l is length of arc)

V∝l

(∵l=rθ)

V∝θ

Let V1 be Volume occupied by mass m1 and V2 be Volume occupied by mass m2

V2V1=θ2θ1

V2V1=α2π−α

Substituting in Equation (i) we get,

V2V1=n2n1

α2π−α=(mm2)(mm1)=m1m2

α2π−α=3228=87

16π−8α=7α

∴α=16π15

Why this question?Tip: In this problem the first crucial information is the conducting nature of movable partition S, which givesinsight that it will enable exchange of heat betweenboth gases to maintain thermal equilibrium i.e. T2=T2Trick : The angle subtended by are will give relation betwenrespective lengths of arc, thus volume of gas in eachsection can be correlated.

At the condition of equilibrium, the pressure of gas on both sides of movable partition S will be equal.

⇒P1=P2

Also the partition S is conducting hence it will maintain the thermal equilibrium of both gases by allowing exchange of heat between then.

Hence at the equilibrium position,

T1=T2

Considering the thermodynamic process to be reversible. because friction is absent,

From ideal gas equation,

P1=P2

n1RT1V1=n2RT2V2

n1V1=n2V2(∵T1=T2).....(i)

Equal mass of both the gases is present inside tube,

n1=mm1& n2=mm2

Also the area of cross - section of tube is constant.

⇒ volume of the gas =A×l

(l is length of arc)

V∝l

(∵l=rθ)

V∝θ

Let V1 be Volume occupied by mass m1 and V2 be Volume occupied by mass m2

V2V1=θ2θ1

V2V1=α2π−α

Substituting in Equation (i) we get,

V2V1=n2n1

α2π−α=(mm2)(mm1)=m1m2

α2π−α=3228=87

16π−8α=7α

∴α=16π15

Why this question?Tip: In this problem the first crucial information is the conducting nature of movable partition S, which givesinsight that it will enable exchange of heat betweenboth gases to maintain thermal equilibrium i.e. T2=T2Trick : The angle subtended by are will give relation betwenrespective lengths of arc, thus volume of gas in eachsection can be correlated.

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