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  • Figure shown a vertical cylindrical vessel separated in two parts by a frictionless pistion free to move along the length of the vessel. The length of the cylinder is <img alt="90 \mathrm{~cm}" src

Figure shown a vertical cylindrical vessel separated in two parts by a frictionless pistion free to move along the length of the vessel. The length of the cylinder is 90 \mathrm{~cm} and the pistion 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 \mathrm{~K} in each part. Calculate the mass of the piton.

Option: 1

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Option: 2

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Option: 3

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Option: 4

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Answers (1)

best_answer

Let \mathrm{l_{1}} and \mathrm{l_{2}}  be the lengths of the upper part and the lower part of the cylinder respectively. clearly, \mathrm{l_{1}=50 \mathrm{~cm}} and \mathrm{l_{2}=40 \mathrm{~cm}}. Let the perssures in the upper and lower parts be \mathrm{p}_{1} and \mathrm{p}_{2} respectively. Let the area of coss-section of the cylinder be \mathrm{A}. The temperature in both parts is \mathrm{\mathrm{T}=300 \mathrm{~K}}.

Consider the equilibrium of the piston. The forces acting on the pistion are
(a) its weight mg
(b) \mathrm{p_{1}}  A downward, by the upper part of the gas and (c) \mathrm{p_{2}A} upward, by the lower part of the gas.
Thus, \mathrm{ \mathrm{p}_{2} \mathrm{~A}=\mathrm{p}_{1} \mathrm{~A}+\mathrm{mg}\quad \ldots(i)}

 Using \mathrm{pV}=\mathrm{nRT}for the upper and the lower parts
\mathrm{\mathrm{p}_{1} l_{1} \mathrm{~A}=\mathrm{nRT} \ldots \text { (ii) }}
\mathrm{\text { and } \mathrm{p}_{2} l_{2} \mathrm{~A}=\mathrm{nRT} \ldots \text { (iii) } }

Putting \mathrm{p_{1} A } and \mathrm{p_{2} A } from (ii) and (iii) into (i),
\frac{\mathrm{nRT}}{l_{2}}=\frac{\mathrm{nRT}}{l_{1}}+\mathrm{mg}
 

Thus,

\mathrm{m}=\frac{\mathrm{nRT}}{\mathrm{g}}\left[\frac{1}{l_{1}}-\frac{1}{l_{2}}\right]
=\frac{(0.1 \mathrm{~mol})(8.3 \mathrm{~J} / \mathrm{mol}-\mathrm{K})(300 \mathrm{~K})}{9.8 \mathrm{~m} / \mathrm{s}^{2}}\left[\frac{1}{0.4 \mathrm{~m}}-\frac{1}{0.5 \mathrm{~m}}\right]
=12.7 \mathrm{~kg}.
 

Posted by

Riya

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