The air tight and smooth pistons of a cylindrical vessel are connected with a string as shown- Initially pressure and temperature of the gas are P - and T - The atmospheric pressure is also P - At a later time- tension in the string is 38 P - A- where A is the cross sectional area of the cylinder- At this time- theB- 34 T -C- 118 T -D- 138 T -

The correct option is C 118T°T nbsp;+ nbsp;P°A nbsp;= nbsp;PAT nbsp;= nbsp;(P nbsp; nbsp;P°)A nbsp;= nbsp;38P°A nbsp;(given) P nbsp; ...

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Byju's Answer

Standard XII

Physics

Gas laws from KTG

The air tight...

Question

The air tight and smooth pistons of a cylindrical vessel are connected with a string as shown. Initially pressure and temperature of the gas are $P_{\circ}$ and $T_{\circ}$ . The atmospheric pressure is also $P_{\circ}$ . At a later time, tension in the string is $\frac{3}{8} P_{\circ} A$ , where $A$ is the cross sectional area of the cylinder. At this time, the temperature of the gas has become

\frac{3}{8} T_{\circ}

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\frac{3}{4} T_{\circ}

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\frac{11}{8} T_{\circ}

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\frac{13}{8} T_{\circ}

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Solution

The correct option is C $\frac{11}{8} T_{\circ}$
$T + P_{\circ} A = P A T = (P - P_{\circ}) A = \frac{3}{8} P_{\circ} A (given)$

$P = \frac{3}{8} P_{\circ} + P_{\circ} = \frac{11}{8} P_{\circ}$
Now volume is constant.
So, $P \propto T$
Initially temperature is $T_{0}$
Now as pressure has become $\frac{11}{8} P_{0}$ , temperature will be $\frac{11}{8} T_{0}$

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The correct option is C 118T∘T + P∘A = PAT = (P − P∘)A = 38P∘A (given) P = 38P∘ + P∘ = 118P∘ Now volume is constant. So, P∝T Initially temperature is T0 Now as pressure has become 118P0, temperature will be 118T0