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Standard XII

Chemistry

Charles' Law

A cylindrical...

Question

A cylindrical tube of uniform cross-sectional area A is filled with two air tight frictionless pistons. The pistons are connected to each other by a metallic wire. Initially, the pressure of the gas is $p_{o}$ and temperature is $T_{o}$ , atmospheric pressure is also $p_{o}$ $. N o w, t h e t e m p e r a t u r e o f t h e g a s i s i n c r e a s e d t o$ 2T_0$, the tension of wire will be.

2 p_{o} A

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p_{o} A

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\frac{p_{o} A}{2}

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4 p_{o} A

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Solution

The correct option is A $p_{o} A$
Volume of the gas is constant i.e., $V =$ constant
$∴ p \propto T$ i.e., pressure will be doubled if temperature is doubled.
Let F be the tension in the wire. Then, equilibrium of any one pipes gives.
$F = (p - p_{0}) A = (2 p_{0} - p_{0}) A = p_{0} A$ .

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The correct option is A poAVolume of the gas is constant i.e., V= constant∴p∝T i.e., pressure will be doubled if temperature is doubled.Let F be the tension in the wire. Then, equilibrium of any one pipes gives.F=(p−p0)A=(2p0−p0)A=p0A.

The correct option is A $p_{o} A$
Volume of the gas is constant i.e., $V =$ constant
$∴ p \propto T$ i.e., pressure will be doubled if temperature is doubled.
Let F be the tension in the wire. Then, equilibrium of any one pipes gives.
$F = (p - p_{0}) A = (2 p_{0} - p_{0}) A = p_{0} A$ .