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Byju's Answer

Standard XII

Chemistry

Charles' Law

A cylindrical...

Question

A cylindrical tube of cross-sectional area $A$ has two air tight frictionless pistons at its two ends. The pistons are tied with a straight two ends. The pistons are tied with a straight piece of metallic wire. The tube contains a gas at atmospheric pressure $P_{0}$ and temperature $T_{0}$ . If temperature of the gas is doubled then the tension in the wire is

4 P_{0} A

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P_{0} A / 2

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P_{0} A

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2 P_{0} A

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Solution

The correct option is C $P_{0} A$
The volume of the gas is constant
i.e., $v =$ constant.
$∴ P \propto T$
Pressure will be doubled if the temeprature is doubled.
Let $F$ be the tension in the wire. Then the equilibrium of any one pipe gives
$F = (P - P_{0}) A = (2 P_{0} - P_{0}) A$
$F = P_{0} A$

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The correct option is C P0AThe volume of the gas is constant i.e., v= constant. ∴P∝T Pressure will be doubled if the temeprature is doubled. Let F be the tension in the wire. Then the equilibrium of any one pipe gives F=(P−P0)A=(2P0−P0)A F=P0A

The correct option is C $P_{0} A$
The volume of the gas is constant
i.e., $v =$ constant.
$∴ P \propto T$
Pressure will be doubled if the temeprature is doubled.
Let $F$ be the tension in the wire. Then the equilibrium of any one pipe gives
$F = (P - P_{0}) A = (2 P_{0} - P_{0}) A$
$F = P_{0} A$