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Question

Figure shows a vertical cylindrical vessel separated in two parts by a frictionless piston free to move along the length of the vessel. The length of the cylinder is 90 cm and the piston divides the cylinder in the ratio of 5:4. Each of the two parts of the vessel contains 0.1 mole of an ideal gas. The temperature of the gas is 300 K in each part. Calculate the mass of the piston


A

10.3 kg

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B

5.7 kg

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C

12.7 kg

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D

15.5 kg

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Solution

The correct option is C

12.7 kg


Let l1 and l2 be the lengths of the upper part and the lower part of the cylinder respectively. Clearly, l1=50cm and l2=40cm. Let the pressures in the upper and lower parts be P1 and P2 respectively. Let the area of cross section of the cylinder be A. The temperature in both parts is T=300K.

Consider the equilibrium of the piston. The forces acting on the piston are

(a) its weight mg

(b) P1A downward, by the upper part of the gas and

(c) P2A upward, by the lower part of the gas.

Thus,

P2A=P1A+mg. .....(i)

Using PV = nRT for the upper and the lower parts

P1l1A=nRT .....(ii)

and P2l2A=nRT ....(iii)

Putting P1A and P2A from (ii) and (iii) into (i),

nRTl2=nRTl1+mg.

Thus,

m=nRTg[1l21l1]

=[(0.1 mol)×(8.3JK1mol1)×(300 K)9.8 m s2][10.4 m10.5 m]=12.7 kg.

Do you see the similarities with the previous problem, where we had the mercury column?

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