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Standard XII

Physics

Conduction Law

A monoatomic ...

Question

A monoatomic ideal gas is contained in a rigid container of volume $V$ with walls of total surface area $A$ , thickness $x$ and thermal conductivity $K$ . The gas is at an initial temperature $T_{o}$ and pressure $P_{o}$ . The temperature of the surrounding air is $T$ . All the temperatures are in absolute scale. If the temperature of the gas at any time $t$ can be expressed as $T (t) = T_{s} - (T_{s} - T_{o}) e^{- λ t}$ , find the pressure of the gas at the steady state condition.

P = P_{o}

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P = 2 P_{o}

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P = \frac{P_{o}}{T_{o}} T_{s}

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None of the above

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Solution

The correct option is C $P = \frac{P_{o}}{T_{o}} T_{s}$
As the process is isochoric, $\frac{P_{o}}{T_{o}} = \frac{P (t)}{T (t)}$
[fixed volume of container]
$∵ T (t) = T_{s} - (T_{s} - T_{o}) e^{- λ t}$
$P (t) = \frac{P_{o}}{T_{o}} [T_{s} - (T_{s} - T_{o}) e^{- λ t}]$
At steady state, as $t \to \infty$ , second term goes to zero.
$\Rightarrow P = \frac{P_{o}}{T_{o}} T_{s}$

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The correct option is C P=PoToTsAs the process is isochoric, PoTo=P(t)T(t) [fixed volume of container] ∵T(t)=Ts−(Ts−To)e−λt P(t)=PoTo[Ts−(Ts−To)e−λt] At steady state, as t→∞, second term goes to zero. ⇒P=PoToTs