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Standard XII

Physics

Ideal Gas Equation

A cylindrical...

Question

A cylindrical tube of cross sectional area A has two air tight frictionless pistons at its two ends. The piston 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 the 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 $P_{0} A$
PV = nRT
In this case V is constant.
So doubling T also doubles P.
Tension = Pressure x Area
net pressure=gas pressure-atmospheric pressure
$P_{n e t} = 2 P - P = P$
Therefore, new Tension = PA

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The correct option is C P0APV = nRTIn this case V is constant. So doubling T also doubles P.Tension = Pressure x Areanet pressure=gas pressure-atmospheric pressurePnet=2P−P=PTherefore, new Tension = PA

The correct option is C $P_{0} A$
PV = nRT
In this case V is constant.
So doubling T also doubles P.
Tension = Pressure x Area
net pressure=gas pressure-atmospheric pressure
$P_{n e t} = 2 P - P = P$
Therefore, new Tension = PA