11.Thermodynamics
medium

एक आदर्श गैस को $27°C$ पर रुद्धोष्म संपीडित किया जाता है कि उसका आयतन, प्रारम्भिक आयतन का $8/27 $ गुना हो जाता है यदि $\gamma = 5/3$ है, तो ताप वृद्धि ....... $K$ होगी

A

$450 $

B

$375 $

C

$225 $

D

$405 $

(AIPMT-1999)

Solution

$\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^{\gamma  – 1}} \Rightarrow {T_2} = 300\,{\left( {\frac{{27}}{8}} \right)^{\frac{5}{3} – 1}} = 300\,{\left( {\frac{{27}}{8}} \right)^{\frac{2}{3}}}$

$ = 300\,{\left\{ {{{\left( {\frac{{27}}{8}} \right)}^{1/3}}} \right\}^2} = 800\,{\left( {\frac{3}{2}} \right)^2} = 675K$

$ \Rightarrow \;\Delta T = 675 – 300 = 375\;K$

Standard 11
Physics

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